Arithmetic & Geometric Progression

(i) First term = a = 3 and common difference = d = 1 - 3 = -2.

(ii) First term = a = $\frac{1}{3}$ and common difference = d = $\frac{5}{3} - \frac{1}{3} = \frac{4}{3}$.

(iii) First term = a = -3.2 and common difference = d = -3 - (-3.2) = 0.2.

(i) Here, a₁ = a = 10,   a₂ = a₁ + d = 10 + 10 = 20,

a₃ = a₂ + d = 20 + 10 = 30,    a₄ = a₃ + d = 30 + 10 = 40.

Hence, the first four terms of A.P. are 10, 20, 30, 40.

(ii) Here, a₁ = a = -2,   a₂ = a₁ + d = -2 + 0 = -2,

a₃ = a₂ + d = -2 + 0 = -2,    a₄ = a₃ + d = -2 + 0 = -2.

Hence, the first four terms of A.P. are -2, -2, -2, -2.

(iii) Here, a₁ = a = 4,   a₂ = a₁ + d = 4 + (-3) = 1,

a₃ = a₂ + d = 1 + (-3) = -2,    a₄ = a₃ + d = -2 + (-3) = -5.

Hence, the first four terms of A.P. are 4, 1, -2, -5.

(iv) Here, a₁ = a = $\frac{1}{2}$,   a₂ = a₁ + d = $\frac{1}{2} + \big(-\frac{1}{6}\big) = \frac{2}{6} = \frac{1}{3}$,

a₃ = a₂ + d = $\frac{2}{6} + \big(-\frac{1}{6}\big) = \frac{1}{6}$,    a₄ = a₃ + d = $\frac{1}{6} + \big(-\frac{1}{6}\big)$ = 0.

Hence, the first four terms of A.P. are $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}, 0$.

(i) Given, 4, 10, 16, 22, ....

Here, a₂ - a₁ = 10 - 4 = 6,     a₃ - a₂ = 16 - 10 = 6,

          a₄ - a₃ = 22 - 16 = 6

i.e. any term - preceding term = 6, a fixed number.
Hence, the given list of numbers forms an A.P. with common difference = d = 6.

For the next three terms, we have:
    a₅ = a₄ + d = 22 + 6 = 28,
    a₆ = a₅ + d = 28 + 6 = 34,
    a₇ = a₆ + d = 34 + 6 = 40.

Hence, the given series is in A.P. with common difference d = 6 and the next three terms : 28, 34, 40.

(ii) Given, -2, 2, -2, 2, ....

Here, a₂ - a₁ = 2 - (-2) = 4,     a₃ - a₂ = -2 - 2 = -4,

          a₄ - a₃ = 2 - (-2) = 4

⇒ a₂ - a₁ = a₄ - a₃ ≠ a₃ - a₂.

Thus the difference of any term from its preceding term is not a fixed number.

Hence, the given series does not form an A.P.

(iii) Given, 2, 4, 8, 16, ....

Here, a₂ - a₁ = 4 - 2 = 2,     a₃ - a₂ = 8 - 4 = 4,

          a₄ - a₃ = 16 - 8 = 8

⇒ a₂ - a₁ ≠ a₃ - a₂ ≠ a₄ - a₃.

Thus the difference of any term from its preceding term is not a fixed number.

Hence, the given series does not form an A.P.

(iv) Given, $2, \frac{5}{2}, 3, \frac{7}{2}, ....$

Here, a₂ - a₁ = $\frac{5}{2} - 2 = \frac{1}{2}$,     a₃ - a₂ = $3 - \frac{5}{2} = \frac{1}{2}$,

          a₄ - a₃ = $\frac{7}{2} - 3 = \frac{1}{2}$

i.e. any term - preceding term = $\frac{1}{2}$, a fixed number.
Hence, the given list of numbers forms an A.P. with common difference = d = $\frac{1}{2}$.

For the next three terms, we have:

    a₅ = a₄ + d = $\frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4$,

    a₆ = a₅ + d = $4 + \frac{1}{2} = \frac{9}{2}$,

    a₇ = a₆ + d = $\frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5$.

Hence, the given series is in A.P. with common difference d = $\frac{1}{2}$ and the next three terms : 4, $\frac{9}{2}$, 5.

(v) Given, -10, -6, -2, 2, ...

Here, a₂ - a₁ = -6 - (-10) = 4,     a₃ - a₂ = -2 - (-6) = 4,

          a₄ - a₃ = 2 - (-2) = 4

i.e. any term - preceding term = 4, a fixed number.
Hence, the given list of numbers forms an A.P. with common difference = d = 4.

For the next three terms, we have:
    a₅ = a₄ + d = 2 + 4 = 6,
    a₆ = a₅ + d = 6 + 4 = 10,
    a₇ = a₆ + d = 10 + 4 = 14.

Hence, the given series is in A.P. with common difference d = 4 and the next three terms : 6, 10, 14.

(vi) Given, 1², 3², 5², 7², ....

or, 1, 9, 25, 49 ....

Here, a₂ - a₁ = 9 - 1 = 8,     a₃ - a₂ = 25 - 9 = 16,

          a₄ - a₃ = 49 - 25 = 24

⇒ a₂ - a₁ ≠ a₃ - a₂ ≠ a₄ - a₃.

Thus the difference of any term from its preceding term is not a fixed number.

Hence, the given series does not form an A.P.

Given, a_n = 7 - 3n     (Eq 1)

Putting n = 1, 2, 3, 4, .... in (Eq 1), we get

a₁ = 7 - 3 × 1 = 7 - 3 = 4,   a₂ = 7 - 3 × 2 = 7 - 6 = 1
a₃ = 7 - 3 × 3 = 7 - 9 = -2,   a₄ = 7 - 3 × 4 = 7 - 12 = -5, ...

∴ Series = 4, 1, -2, -5 .....

Putting n = 20, for 20th term we get,

a₂₀ = 7 - 3 × 20 = 7 - 60 = -53.

Hence, the A.P. is 4, 1, -2, -5 ..... and the 20th term is -53.

(i) The given list of numbers is 1, 6, 11, 16, ...

Here, a₂ - a₁ = 6 - 1 = 5,     a₃ - a₂ = 11 - 6 = 5,

          a₄ - a₃ = 16 - 11 = 5

i.e. any term - preceding term = 5, a fixed number.
So, the given list of numbers forms an A.P. with a = 1 and d = 5.

∴ nth term = a_n = a + (n - 1)d = 1 + (n - 1)5 = 1 + 5n - 5 = 5n - 4.

Putting n = 20, we get
a₂₀ = 5 × 20 - 4 = 100 - 4 = 96.

Hence, a₂₀ = 96.

(ii) The given list of numbers is -4, -7, -10, -13, ....

Here, a₂ - a₁ = -7 - (-4) = -3,     a₃ - a₂ = -10 - (-7) = -3,

          a₄ - a₃ = -13 - (-10) = -3

i.e. any term - preceding term = -3, a fixed number.
So, the given list of numbers forms an A.P. with a = -4 and d = -3.

∴ nth term = a_n = a + (n - 1)d = -4 + (n - 1)(-3) = -4 - 3n + 3 = -3n - 1.

Putting n = 25, we get
a₂₅ = -3 × 25 - 1 = -75 - 1 = -76.

Hence, a₂₅ = -76 and a_n = -3n - 1.

The given list of numbers is 5, 2, -1, -4, ....

Here, a₂ - a₁ = 2 - 5 = -3,     a₃ - a₂ = -1 - 2 = -3,

          a₄ - a₃ = -4 - (-1) = -3

i.e. any term - preceding term = -3, a fixed number.
So, the given list of numbers forms an A.P. with a = 5 and d = -3.

∴ nth term = a_n = a + (n - 1)d = 5 + (n - 1)(-3) = 5 - 3n + 3 = 8 - 3n.

Putting n = 12, we get
a₁₂ = 8 - 3(12) = 8 - 36 = -28.

Hence, a₁₂ = -28 and a_n = 8 - 3n.

The nth term of an A.P. is given by,

a_n = a + (n - 1)d.    (Eq 1)

Given , d = -3 and a₁₈ = -5.

Putting d = -3, a₁₈ = -5 and n = 18 in Eq 1 we get,

⇒ a_n = a + (n - 1)d
⇒ -5 = a + (18 - 1)(-3)
⇒ -5 = a + 17 × (-3)
⇒ -5 = a - 51
⇒ a = 51 - 5
⇒ a = 46.

Hence, the first term of A.P. is 46.

The nth term of an A.P. is given by,

a_n = a + (n - 1)d.    (Eq 1)

Given , a = -18 and a₁₀ = 0.

Putting a = -18, a₁₀ = 0 and n = 10 in Eq 1 we get,

⇒ a_n = a + (n - 1)d
⇒ 0 = -18 + (10 - 1)d
⇒ 0 = -18 + 9d
⇒ 9d = 18
⇒ d = 2.

Hence, the common difference of A.P. is 2.

(i) The given A.P. is 3, 8, 13, 18 ....

Here, first term = a = 3 and common difference = d = 8 - 3 = 5.

Let 78 be nth term of the A.P., or a_n = 78

The nth term of A.P. is given by a_n = a + (n - 1)d

Putting value of d and a_n in above equation we get,

⇒ a_n = a + (n - 1)d
⇒ 78 = 3 + (n - 1)5
⇒ 78 = 3 + 5n - 5
⇒ 78 = 5n - 2
⇒ 5n = 78 + 2
⇒ 5n = 80
⇒ n = 16.

Hence, 78 is 16th term of the A.P.

(ii) The given A.P. is $18, 15\frac{1}{2}, 13, ...$

Here, first term = a = 18 and common difference = d = $\frac{31}{2} - 18 = \frac{31 - 36}{2} = -\frac{5}{2}$.

Let -47 be nth term of the A.P., or a_n = -47

The nth term of A.P. is given by a_n = a + (n - 1)d

Putting value of a, d and a_n in above equation we get,

$$\begin{aligned}a_n = a + (n - 1)d \\[1em] \Rightarrow -47 = 18 + (n - 1)\Big(-\frac{5}{2}\Big) \\[1em] \Rightarrow -47 = 18 -\frac{5n}{2} + \frac{5}{2} \\[1em] \Rightarrow -47 - 18 = \frac{5 - 5n}{2} \\[1em] \Rightarrow -65 \times 2 = 5 - 5n \\[1em] \Rightarrow -130 = 5 - 5n \\[1em] \Rightarrow 5n = 130 + 5 \\[1em] \Rightarrow 5n = 135 \\[1em] \Rightarrow n = 27.\end{aligned}$$

Hence, -47 is 27th term of the A.P.

Here, a₂ - a₁ = 8 - 11 = -3,     a₃ - a₂ = 5 - 8 = -3,

          a₄ - a₃ = 2 - 5 = -3

i.e. any term - preceding term = -3, a fixed number.
So, the given list of numbers forms an A.P. with a = 11 and d = -3.

Now we want to find whether there exists a natural number n for which a_n = -150

⇒ a + (n - 1)d = -150
⇒ 11 + (n - 1)(-3) = -150
⇒ 11 - 3n + 3 = -150
⇒ 14 - 3n = -150
⇒ 3n = 14 + 150
⇒ 3n = 164
⇒ n = $54\frac{2}{3}$, which is not a natural number.

Hence, there is no term in the given list of numbers which is -150.

Here, a₂ - a₁ = 10 - 7 = 3,     a₃ - a₂ = 13 - 10 = 3,

i.e. any term - preceding term = 3, a fixed number.
So, the given list of numbers forms an A.P. with a = 7 and d = 3.

Now we want to find whether there exists a natural number n for which a_n = 55

⇒ a + (n - 1)d = 55
⇒ 7 + (n - 1)3 = 55
⇒ 7 + 3n - 3 = 55
⇒ 4 + 3n = 55
⇒ 3n = 51
⇒ n = 17.

Hence, 55 is the 17th term of the A.P. 7, 10, 13 .....

Here, common difference = d = 8 - 3 = 5 and last term = l = 253.

We know that the nth term from last is given by the formula:
nth term from end = l - (n - 1)d

∴ 20th term from end = 253 - (20 - 1)5 = 253 - 19 × 5 = 253 - 95 = 158.

Hence, the 20th term from last term of the A.P. 3, 8, 13, ...., 253 is 158.

Here, common difference = d = -4 - (-2) = -2 and last term = l = -100.

We know that the nth term from last is given by the formula:
nth term from end = l - (n - 1)d

∴ 20th term from end = -100 - (12 - 1)(-2) = -100 - 11 × (-2) = -100 + 22 = -78.

Hence, the 12th term from last term of the A.P. -2, -4, -6, ...., -100 is -78.

Here, a = $-\frac{4}{3} \text{ and d } = -1- \big(-\frac{4}{3}\big) = \frac{-3 + 4}{3} = \frac{1}{3}$

$\text{ and l } = 4\frac{1}{3} = \frac{13}{3}.$

We need to find the number of terms in the A.P.

$$\begin{aligned}\Rightarrow a_n = a + (n - 1)d \\[1em] \Rightarrow \frac{13}{3} = -\frac{4}{3} + (n - 1)\frac{1}{3} \\[1em] \Rightarrow \frac{13}{3} + \frac{4}{3} = \frac{(n - 1)}{3} \\[1em] \Rightarrow \frac{17}{3} = \frac{(n - 1)}{3}\end{aligned}$$

On multiplying both sides by 3,

$$\begin{aligned}\Rightarrow 17 = n - 1 \\[1em] \Rightarrow 17 + 1 = n \\[1em] \Rightarrow n = 18.\end{aligned}$$

Since, A.P. has 18 terms, therefore 9th and 10th terms are two middle most terms.

a₉ = a + (9 - 1)d = a + 8d        (Eq 1)

a₁₀ = a + (10 - 1)d = a + 9d     (Eq 2)

Adding Eq 1 and Eq 2,

a₉ + a₁₀ = a + 8d + a + 9d = 2a + 17d.

Hence, the sum of middle most terms

$$\begin{aligned}= 2a + 17d \\[0.5em] = 2 \times -\frac{4}{3} + 17 \times \frac{1}{3} \\[0.5em] = -\frac{8}{3} + \frac{17}{3} \\[0.5em] = \frac{-8 + 17}{3} \\[0.5em] = \frac{9}{3} \\[0.5em] = 3.\end{aligned}$$

Hence, the sum of the two middle most terms of the A.P. is equal to 3.

Let us first evaluate which term is 0 because the next term will be the term containing first negative number.

a = 53, d = 48 - 53 = -5, a_n = 0.

By formula, a_n = a + (n - 1)d
⇒ a_n = 53 + (n - 1)(-5)
⇒ 0 = 53 - 5n + 5
⇒ 5n = 58
⇒ n = $11\frac{3}{5}$

The next term will be containing the first negative number.

Hence, the 12th term will contain the first negative number.

Given, a₃ = 16 (Eq 1)     and    a₇ - a₅ = 12 (Eq 2)

By using formula a_n = a + (n - 1)d on Eq 1 we get,

⇒ a₃ = a + (3 - 1)d = 16
⇒ a + 2d = 16
⇒ a = 16 - 2d.             (Eq 3)

By using formula a_n = a + (n - 1)d on Eq 2 we get,

⇒ a₇ - a₅ = 12
⇒ a + (7 - 1)d - [a + (5 - 1)d] = 12
⇒ a + 6d - a - 4d = 12
⇒ 2d = 12 ⇒ d = 6.

∴ a = 16 - 2d = 16 - 2(6) = 16 - 12 = 4.

    a₂ = a + d = 4 + 6 = 10,    a₃ = a₂ + d = 10 + 6 = 16,
    a₄ = a₃ + d = 16 + 6 = 22.

Hence, the A.P. is 4, 10, 16, 22 ....

Given, a = 12 and a₁₁ - a₇ = 24 (Eq 1)

By using formula a_n = a + (n - 1)d for Eq 1 we get,

⇒ a₁₁ - a₇ = 24
⇒ a + (11 - 1)d - [a + (7 - 1)d] = 24
⇒ a + 10d - a - 6d = 24
⇒ 4d = 24
⇒ d = 6.

20th term of A.P. is,

a₂₀ = 12 + (20 - 1)6 = 12 + 19 × 6 = 12 + 114 = 126.

Hence, the 20th term of the A.P. is 126.

Given,
a₁₁ = 38     (Eq 1)
a₆ = 73       (Eq 2)

By using formula a_n = a + (n - 1)d for Eq 1 we get,

⇒ a₁₁ = a + (11 - 1)d = 38
⇒ a + 10d = 38
⇒ a = 38 - 10d              (Eq 3)

By using formula a_n = a + (n - 1)d for Eq 2 we get,

⇒ a₆ = a + (6 - 1)d = 73
⇒ a + 5d = 73                (Eq 4)

Putting value of a from Eq 3 in Eq 4 we get,

⇒ a + 5d = 73
⇒ 38 - 10d + 5d = 73
⇒ 38 - 5d = 73
⇒ 5d = 38 - 73
⇒ 5d = -35
⇒ d = -7.

∴ a = 38 - 10d = 38 - 10(-7) = 38 + 70 = 108.

and 31st term = a₃₁ = a + (31 - 1)d = 108 + 30(-7) = 108 - 210 = -102.

Hence, the 31st term of the A.P. is -102.

Given,

a₇ = $\frac{1}{9}$     (Eq 1)

a₉ = $\frac{1}{7}$     (Eq 2)

By using formula a_n = a + (n - 1)d for Eq 1 we get,

⇒ a₇ = a + (7 - 1)d = $\frac{1}{9}$

⇒ a + 6d = $\frac{1}{9}$

⇒ 9(a + 6d) = 1

⇒ 9a + 54d = 1

⇒ 9a = 1 - 54d

⇒ a = $\frac{1 - 54d}{9}$     (Eq 3)

By using formula a_n = a + (n - 1)d for Eq 2 we get,

⇒ a₉ = a + (9 - 1)d = $\frac{1}{7}$

⇒ a + 8d = $\frac{1}{7}$     (Eq 4)

Putting value of a from Eq 3 in Eq 4 above,

$$\begin{aligned}\Rightarrow \frac{1 - 54d}{9} + 8d = \frac{1}{7} \\[1em] \Rightarrow \frac{1 - 54d + 72d}{9} = \frac{1}{7} \\[1em] \Rightarrow \frac{1 + 18d}{9} = \frac{1}{7} \\[1em] \Rightarrow 7 + 126d = 9 \\[1em] \Rightarrow 126d = 2 \\[1em] \Rightarrow d = \frac{2}{126} \\[1em] \Rightarrow d = \frac{1}{63} \\[1.5em] \therefore a = \frac{1 - 54d}{9} \\[1em] = \frac{1 - 54 \times \frac{1}{63}}{9} \\[1em] = \frac{63 - 54}{9 \times 63} \\[1em] = \frac{9}{9 \times 63} \\[1em] = \frac{1}{63}. \\[1em]\end{aligned}$$

63rd term = a₆₃ = $a + (63 - 1)d$

$$\begin{aligned}= a + 62d \\[1em] = \frac{1}{63} + 62 \times \frac{1}{63} \\[1em] = \frac{1}{63} + \frac{62}{63} \\[1em] = \frac{63}{63} \\[1em] = 1.\end{aligned}$$

Hence, 63rd term of the A.P. is 1.

Given,

a₁₀ = 41,          (Eq 1)
a₁₅ = 2a₇ + 3.  (Eq 2)

By using formula a_n = a + (n - 1)d for Eq 1 we get,
⇒ a₁₀ = a + (10 - 1)d = 41
⇒ a + 9d = 41
⇒ a = 41 - 9d     (Eq 3)

By using formula a_n = a + (n - 1)d for Eq 2 we get,
⇒ a₁₅ = 2a₇ + 3
⇒ a + (15 - 1)d = 2(a + (7 - 1)d) + 3
⇒ a + 14d = 2(a + 6d) + 3
⇒ a + 14d = 2a + 12d + 3

Putting value of a from Eq 3 in above equation
⇒ 41 - 9d + 14d = 2(41- 9d) + 12d + 3
⇒ 41 + 5d = 82 - 18d + 12d + 3
⇒ 41 + 5d = 82 - 6d + 3
⇒ 5d + 6d = 85 - 41
⇒ 11d = 44
⇒ d = 4.

∴ a = 41 - 9d = 41 - 9(4) = 41 - 36 = 5.

nth term = a_n = a + (n - 1)d = 5 + 4(n - 1) = 5 + 4n - 4 = 4n + 1.

Hence, the nth term of the A.P. is 4n + 1.

Given,

a₅ + a₇ = 52   (Eq 1)

a₁₀ = 46         (Eq 2)

By using formula a_n = a + (n - 1)d for Eq 1 we get,
⇒ a₅ + a₇ = 52
⇒ a + (5 - 1)d + a + (7 - 1)d = 52
⇒ a + 4d + a + 6d = 52
⇒ 2a + 10d = 52
⇒ 2(a + 5d) = 52
⇒ a + 5d = 26
⇒ a = 26 - 5d    (Eq 3)

By using formula a_n = a + (n - 1)d for Eq 2 we get,
⇒ a₁₀ = 46
⇒ a + (10 - 1)d = 46
⇒ a + 9d = 46

Putting value of a from Eq 3 in above equation
⇒ 26 - 5d + 9d = 46
⇒ 26 + 4d = 46
⇒ 4d = 46 - 26
⇒ 4d = 20
⇒ d = 5.

∴ a = 26 - 5d = 26 - 5 × 5 = 26 - 25 = 1.

We know a₁ = a so,

a₂ = a₁ + d = 1 + 5 = 6,
a₃ = a₂ + d = 6 + 5 = 11,
a₄ = a₃ + d = 11 + 5 = 16.

Hence, the required A.P. is 1, 6, 11, 16, 21, ...

Given,
a₈ = 0

By using formula a_n = a + (n - 1)d for a₈ we get,

⇒ a + (8 - 1)d = 0
⇒ a + 7d = 0
⇒ a = -7d.

So,
a₃₈ = a + (38 - 1)d
      = a + 37d
      = -7d + 37d
      = 30d.

a₁₈ = a + (18 - 1)d
      = a + 17d
      = -7d + 17d
      = 10d.

∴ a₃₈ = 30d = 3 × 10d = 3 × a₁₈.

Hence, proved that 38th term of the A.P. is triple of the 18th term.

Here, common difference = d = 10 - 3 = 7 and a = 3.

Let the nth term of A.P. be 84 more than its 13th term so,

⇒ a_n - a₁₃ = 84
⇒ a + (n - 1)d - (a + (13 - 1)d) = 84
⇒ 3 + 7(n - 1) - (3 + 12 × 7) = 84
⇒ 3 + 7n - 7 - (3 + 84) = 84
⇒ 7n - 4 - 87 = 84
⇒ 7n - 91 = 84
⇒ 7n = 84 + 91
⇒ 7n = 175
⇒ n = 25.

Hence, the 25th term of A.P. will be 84 more than 13th term.

Two digit numbers which are divisible by 3 are :

12, 15, 18, 21, 24, ......, 99.

First number = a = 12 and common difference = d = 15 - 12 = 3.

Last number = 99.

By formula, a_n = a + (n - 1)d

⇒ 99 = 12 + 3(n - 1)
⇒ 99 = 12 + 3n - 3
⇒ 99 = 9 + 3n
⇒ 99 - 9 = 3n
⇒ 3n = 90
⇒ n = 30.

Hence, there are 30 two digit numbers that are divisible by 3.

The numbers which are divisible by both 2 and 5 are
110, 120, 130, 140 .....990.

The above numbers are in A.P. with common difference = d = 120 - 110 = 10 and first term = a = 110.

Last term = 990.

By formula, a_n = a + (n - 1)d

⇒ 990 = 110 + 10(n - 1)
⇒ 990 = 110 + 10n - 10
⇒ 990 = 100 + 10n
⇒ 990 - 100 = 10n
⇒ 10n = 890
⇒ n = 89.

Hence, there are 89 natural numbers between 101 and 999 that are divisible by 2 and 5.

Here, a = a₁ = n - 2, a₂ = 4n - 1 and a₃ = 5n + 2

Since, the numbers are in A.P. so,
a₂ - a₁ = d = a₃ - a₂

⇒ a₂ - a₁ = a₃ - a₂
⇒ 4n - 1 - (n - 2) = 5n + 2 - (4n - 1)
⇒ 4n - n - 1 + 2 = 5n - 4n + 2 + 1
⇒ 3n + 1 = n + 3
⇒ 3n - n = 3 - 1
⇒ 2n = 2
⇒ n = 1.

Hence, the value of n = 1.

Let the three numbers that are in A.P. be a - d, a, a + d.

Given, sum of three numbers = 3
⇒ a - d + a + a + d = 3
⇒ 3a = 3
⇒ a = 1.

Given, product of the numbers = -35
⇒ (a - d)(a)(a + d) = -35

Putting the value of a = 1 in the above eq we get,

⇒ (1 - d)(1)(1 + d) = -35
⇒ (1 - d²) = -35
⇒ d² = 1 + 35
⇒ d² = 36
⇒ d = $\sqrt{36}$
⇒ d = +6, -6.

∴ a - d = 1 - 6 = -5 or 1 - (-6) = 7 and a + d = 1 + 6 = 7 or 1 + (-6) = -5

Hence, the numbers are -5, 1, 7.

Let the three numbers that are in A.P. be a - d, a, a + d.

Given, sum of three numbers = 30
⇒ a - d + a + a + d = 30
⇒ 3a = 30
⇒ a = 10.

Given, the ratio of the first number to the third number = 3 : 7

$$\begin{aligned}\Rightarrow \frac{a - d}{a + d} = \frac{3}{7} \\[0.5em] \Rightarrow 7(a - d) = 3(a + d) \\[0.5em] \Rightarrow 7a - 7d = 3a + 3d \\[0.5em] \Rightarrow 7a - 3a = 3d + 7d \\[0.5em] \Rightarrow 4a = 10d \\[0.5em] \Rightarrow d = \frac{4a}{10} \\[0.5em]\end{aligned}$$

Putting the value of a = 10 we get,

$$\begin{aligned}\Rightarrow d = \frac{4 \times 10}{10} \\[0.5em] \Rightarrow d = 4.\end{aligned}$$

∴ a - d = 10 - 4 = 6 and a + d = 10 + 4 = 14.

Hence, the numbers are 6, 10, 14.

Let the three numbers be a - d, a, a + d.

Given , sum of these terms = 33
⇒ a - d + a + a + d = 33
⇒ 3a = 33
⇒ a = 11.

Given, the product of the first and the third terms exceeds the second term by 29
⇒ (a + d)(a - d) - a = 29

Putting the value of a = 11 we get,

⇒ (11 + d)(11 - d) - 11 = 29
⇒ (11)² - d² - 11 = 29
⇒ 121 - d² - 11 = 29
⇒ 110 - d² = 29
⇒ 110 - 29 = d²
⇒ d² = 81
⇒ d = $\sqrt{81}$
⇒ d = +9, -9.

∴ a - d = 11 - 9 = 2 or 11 - (-9) = 20 and a + d = 11 + 9 = 20 or 11 + (-9) = 2.

Hence, the A.P. is 2, 11, 20, ... or 20, 11, 2, ....

(i) Here, a = 2, d = 7 - 2 = 5 and n = 10.

We know that Sum = $\frac{n}{2}[2a + (n - 1)d]$

$$\begin{aligned}\therefore \text{ Sum } = \frac{10}{2}[2 \times 2 + (10 - 1)5] \\[1em] = 5[4 + 9 \times 5] \\[1em] = 5 \times 49 \\[1em] = 245.\end{aligned}$$

Hence, the sum of the A.P. 2, 7, 12, ... upto 10 terms is 245.

(ii) Here, a = $\frac{1}{15}, d = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}$ and n = 11.

We know that Sum = $\frac{n}{2}[2a + (n - 1)d]$

$$\begin{aligned}\therefore \text{ Sum } = \frac{11}{2}[2 \times \frac{1}{15} + (11 - 1)\frac{1}{60}] \\[1em] = \frac{11}{2}\Big[\frac{2}{15} + \frac{10}{60}\Big] \\[1em] = \frac{11}{2} \times \Big[\frac{2 \times 4 + 10}{60}\Big] \\[1em] = \frac{11}{2} \times \frac{18}{60}\end{aligned}$$

Dividing numerator and denominator by 6,

$$\begin{aligned}= \frac{198}{120} \\[1em] = \frac{33}{20}. \\[1em]\end{aligned}$$

Hence, the sum of the A.P. $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, ...$ upto 11 terms is $\frac{33}{20}$.

(i) The given numbers form an A.P. with a = 34 and d = 32 - 34 = -2.

Let nth term be 10,
as, a_n = a + (n - 1)d

∴ 10 = 34 + (n - 1)(-2)
⇒ 10 = 34 - 2n + 2
⇒ 10 = 36 - 2n
⇒ 2n = 36 - 10
⇒ 2n = 26
⇒ n = 13.

$$\begin{aligned}\text{Sum } = \frac{n}{2}[2a + (n - 1)d] \\[1em] \therefore \text{ Sum } = \frac{13}{2}[2 \times 34 + (13 - 1)(-2)] \\[1em] = \frac{13}{2}[68 + 12(-2)] \\[1em] = \frac{13}{2}[68 - 24] \\[1em] = \frac{13}{2} \times 44 \\[1em] = 13 \times 22 \\[1em] = 286.\end{aligned}$$

Hence, the sum of the series 34 + 32 + 30 + .... + 10 is 286.

(ii) The given numbers form an A.P. with a = -5 and d = -8 - (-5) = -3.

Let nth term be -230, then as, a_n = a + (n - 1)d

∴ -230 = -5 + (n - 1)(-3)
⇒ -230 = -5 -3n + 3
⇒ -230 = -2 - 3n
⇒ -230 + 2 = -3n
⇒ -228 = -3n
⇒ 3n = 228
⇒ n = 76.

$$\begin{aligned}\text{Sum } = \frac{n}{2}[2a + (n - 1)d] \\[1 em] \therefore \text{Sum } = \frac{76}{2}[2 \times -5 + (76 - 1)(-3)] \\[1em] = 38[-10 + 75(-3)] \\[1em] = 38[-10 - 225] \\[1em] = 38 \times -235 \\[1em] = -8930.\end{aligned}$$

Hence, the sum of the series -5 + (-8) + (-11) + .... + (-230) is -8930.

(i) a = 5, d = 3, a_n = 50.

By formula a_n = a + (n - 1)d

⇒ 50 = 5 + (n - 1)3
⇒ 50 = 5 + 3n - 3
⇒ 50 = 2 + 3n
⇒ 50 - 2 = 3n
⇒ 48 = 3n
⇒ n = 16.

By formula $S_n = \frac{n}{2}[2a + (n - 1)d]$

$$\begin{aligned}S_n = \frac{16}{2}[2 \times 5 + (16 - 1)3] \\[1em] = 8[10 + 15 \times 3] \\[1em] = 8 \times 55 \\[1em] = 440.\end{aligned}$$

Hence, n = 16 and S_n = S₁₆ = 440.

(ii) a = 7, a₁₃ = 35

By formula a_n = a + (n - 1)d

⇒ 35 = 7 + (13 - 1)d
⇒ 35 = 7 + 12d
⇒ 35 - 7 = 12d
⇒ 28 = 12d
⇒ d = $\frac{28}{12}$ (Dividing by 4)

⇒ d = $\frac{7}{3}.$

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

$$\begin{aligned}S_{13} = \frac{13}{2}[2 \times 7 + (13 - 1)\Big(\frac{7}{3}\Big)] \\[1em] = \frac{13}{2}[14 + 12 \times \frac{7}{3}] \\[1em] = \frac{13}{2} \times [14 + 28] \\[1em] = \frac{13}{2} \times 42 \\[1em] = 13 \times 21 \\[1em] = 273.\end{aligned}$$

Hence, d = $\frac{7}{3}$ and S_n = S₁₃ = 273.

(iii) d = 5, S₉ = 75

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

$$\begin{aligned}S_9 = \frac{9}{2}[2 \times a + (9 - 1)5] \\[1em] \Rightarrow 75 = \frac{9}{2}[2a + 8 \times 5] \\[1em] \Rightarrow 75 = \frac{9}{2} \times [2a + 40] \\[1em] \Rightarrow 75 = 9(a + 20) \\[1em] \Rightarrow 75 = 9a + 180 \\[1em] \Rightarrow 9a = 75 - 180 \\[1em] \Rightarrow 9a = -105 \\[1em] \Rightarrow a = -\frac{105}{9} \\[1em] \Rightarrow a = -\frac{35}{3}\end{aligned}$$

By formula a_n = a + (n - 1)d,

$$\begin{aligned}\Rightarrow a_9 = -\frac{35}{3} + (9 - 1)5 \\[1em] = -\frac{35}{3} + 40 \\[1em] = \frac{-35 + 120}{3} \\[1em] = \frac{85}{3}.\end{aligned}$$

Hence, a = $-\frac{35}{3} \text{ and } a_9 = \frac{85}{3}.$

(iv) a = 8, a_n = 62, S_n = 210.

By formula a_n = a + (n - 1)d

⇒ 62 = 8 + (n - 1)d
⇒ 62 - 8 = (n - 1)d
⇒ (n - 1)d = 54          (Eq 1)

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

$$\begin{aligned}\Rightarrow 210 = \frac{n}{2}[2 \times 8 + (n - 1)d] \\[1em]\end{aligned}$$

Putting value of (n - 1)d from Eq 1 in above equation,

$$\begin{aligned}\Rightarrow \frac{n}{2}[16 + 54] = 210 \\[1em] \Rightarrow \frac{n}{2} \times 70 = 210 \\[1em] \Rightarrow 35n = 210 \\[1em] \Rightarrow n = \frac{210}{35} \\[1em] \Rightarrow n = 6 \\[1em] \text{From Eq 1, } (n - 1)d = 54 \\[1em] \Rightarrow (6 - 1)d = 54 \\[1em] \Rightarrow 5d = 54 \\[1em] \therefore d = \frac{54}{5} \text{ and } n = 6.\end{aligned}$$

Hence, the value of n = 6 and d = $\frac{54}{5}.$

(v) a = 3, n = 8, S = 192.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ 192 = $\frac{8}{2}[2 \times 3 + (8 - 1)d]$
⇒ 192 = 4[6 + 7d]

⇒ $\frac{192}{4}$ = 6 + 7d
⇒ 48 = 6 + 7d
⇒ 7d = 42
⇒ d = 6.

Hence, the value of d is 6.

Let nth be the last term so,
a = 5, l = a_n = 45, S_n = 400.

By formula a_n = a + (n - 1)d
⇒ 45 = 5 + (n - 1)d
⇒ 45 - 5 = (n - 1)d
⇒ (n - 1)d = 40      (Eq 1)

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

$\Rightarrow 400 = \frac{n}{2}[2 \times 5 + (n - 1)d]$

Putting value of (n - 1)d from Eq 1 in above equation,

$$\begin{aligned}\Rightarrow 400 = \frac{n}{2}[10 + 40] \\[1em] \Rightarrow \frac{n}{2} \times 50 = 400 \\[1em] \Rightarrow 25n = 400 \\[1em] \Rightarrow n = 16\end{aligned}$$

Putting value of n in Eq 1 we get,

$$\begin{aligned}\Rightarrow (16 - 1)d = 40 \\[1em] \Rightarrow 15d = 40 \\[1em] \Rightarrow d = \frac{40}{15} = \frac{8}{3}. \\[1em]\end{aligned}$$

Hence, number of terms = 16 and common difference = $\frac{8}{3}.$

Given, a = 15, S₁₅ = 750

By formula, S_n = $\frac{n}{2}[2a + (n - 1)d]$

$$\begin{aligned}\Rightarrow S_{15} = \frac{15}{2}[2 \times 15 + (15 - 1)d] \\[1em] \Rightarrow 750 = \frac{15}{2}[30 + 14d] \\[1em] \Rightarrow 750 \times 2 = 450 + 210d \\[1em] \Rightarrow 450 + 210d = 1500 \\[1em] \Rightarrow 210d = 1500 - 450 \\[1em] \Rightarrow 210d = 1050 \\[1em] \Rightarrow d = \frac{1050}{210} \\[1em] \Rightarrow d = 5.\end{aligned}$$

By formula, a_n = a + (n - 1)d
⇒ a₂₀ = 15 + (20 - 1)5
⇒ a₂₀ = 15 + 95
⇒ a₂₀ = 110.

Hence, the 20th term of the A.P. is 110.

Given, a = 17, l = 350 and d = 9.

Let nth be the last term so,
l = a_n = 350

By formula, a_n = a + (n - 1)d
⇒ 350 = 17 + (n - 1)9
⇒ 350 = 17 + 9n - 9
⇒ 350 = 9n + 8
⇒ 9n = 350 - 8
⇒ 9n = 342
⇒ n = 38.

By formula, S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ S₃₈ = $\frac{38}{2}[2 \times 17 + (38 - 1)9]$
⇒ S₃₈ = 19[34 + 37(9)]
⇒ 19[34 + 333]
⇒ 19(367)
⇒ 6973.

Hence, number of terms = 38 and the sum of the terms = 6973.

The above series is in A.P. because,
any term - preceding term = 3 = common difference.

Let there be n terms so, x = a_n.

Given, a = 1, d = 3 and S_n = 287.

By formula, S_n = $\frac{n}{2}[2a + (n - 1)d]$

$$\begin{aligned}\Rightarrow S_n = \frac{n}{2}[2 \times 1 + (n - 1)3] \\[1em] \Rightarrow 287 = \frac{n}{2}[2 + 3n - 3] \\[1em] \Rightarrow 287 \times 2 = n[3n - 1] \\[1em] \Rightarrow 3n^2 - n = 574 \\[1em] \Rightarrow 3n^2 - n - 574 = 0 \\[1em] 3n^2 - 42n + 41n - 574 = 0 \\[1em] 3n(n - 14) + 41(n - 14) = 0 \\[1em] (3n + 41)(n - 14) = 0 \\[1em] 3n + 41 = 0 \text{ or } n - 14 = 0 \\[1em] n = -\frac{41}{3} \text{ or } n = 14.\end{aligned}$$

Since number of terms cannot be negative so n ≠ $-\frac{41}{3}$

∴ n = 14.

We know x = a_n so,

⇒ x = a + (n - 1)d
⇒ x = 1 + (14 - 1)3
⇒ x = 1 + 13(3)
⇒ x = 1 + 39
⇒ x = 40.

Hence, the value of x = 40.

Let the number of terms required be n.

For the given A.P.,
a = 25, d = 22 - 25 = -3, S_n = 116

By formula, S_n = $\frac{n}{2}[2a + (n - 1)d]$

$$\begin{aligned}\Rightarrow 116 = \frac{n}{2}[2 \times 25 + (n - 1)(-3)] \\[1em] 116 \times 2 = n[50 - 3n + 3] \\[1em] 232 = n[53 - 3n] \\[1em] 232 = 53n - 3n^2 \\[1em] 3n^2 - 53n + 232 = 0 \\[1em] 3n^2 - 24n - 29n + 232 = 0 \\[1em] 3n(n - 8) - 29(n - 8) = 0 \\[1em] (3n - 29)(n - 8) = 0 \\[1em] 3n - 29 = 0 \text{ or } n - 8 = 0 \\[1em] n = \frac{29}{3} \text{ or } n = 8.\end{aligned}$$

Since, number of terms will be a natural number so n ≠ $\frac{29}{3}$

∴ n = 8.

By formula S_n = $\frac{n}{2}[a + l]$

⇒ 116 = $\frac{8}{2}[25 + l]$
⇒ 116 = 4[25 + l]
⇒ 29 = 25 + l
⇒ 29 - 25 = l
⇒ l = 4.

Hence, number of terms = 8 and the last term = 4.

Let numbers of terms be n.
Given, a = 24, d = 21 - 24 = -3 and S_n = 78.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ 78 = $\frac{n}{2}[2 \times 24 + (n - 1)(-3)]$
⇒ 78 × 2 = n[48 - 3n + 3]
⇒ 156 = n[51 - 3n]
⇒ 156 = 51n - 3n²
⇒ 3n² - 51n + 156 = 0
⇒ 3n² - 12n - 39n + 156 = 0
⇒ 3n(n - 4) - 39(n - 4) = 0
⇒ (3n - 39)(n - 4) = 0
⇒ 3n - 39 = 0 or n - 4 = 0
⇒ 3n = 39 or n = 4
⇒ n = 13 or n = 4.

Let's check sum of 4 terms
⇒ a₄ = a₃ + d
⇒ a₄ = 18 + (-3) = 15

∴ S₄ = 24 + 21 + 18 + 15 = 78.

By formula, a_n = a + (n - 1)d
⇒ a₅ = 24 + (5 - 1)(-3)
⇒ a₅ = 24 + 4(-3)
⇒ a₅ = 24 - 12 = 12.

a₆ = a₅ + d = 12 + (-3) = 9
a₇ = a₆ + d = 9 + (-3) = 6
a₈ = a₇ + d = 6 + (-3) = 3
a₉ = a₈ + d = 3 + (-3) = 0
a₁₀ = a₉ + d = 0 + (-3) = -3
a₁₁ = a₁₀ + d = -3 + (-3) = -6
a₁₂ = a₁₁ + d = -6 + (-3) = -9
a₁₃ = a₁₂ + d = -9 + (-3) = -12.

Taking sum of from 5th term to 13th,

12 + 9 + 6 + 3 + 0 - 3 - 6 - 9 - 12 = 0

Since, the sum from 5tn term to 13th is zero hence, it will not add a value.

∴ n = 13 and 4.

Hence, the number of terms that can be taken for sum to be 78 can be 4 or 13.

Given, d = 7 and a₂₂ = 149.

By formula, a_n = a + (n - 1)d

⇒ 149 = a + (22 - 1)7
⇒ 149 = a + 21(7)
⇒ 149 = a + 147
⇒ a = 149 - 147
⇒ a = 2.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ S_n = $\frac{22}{2}[2 \times 2 + (22 - 1)7]$
⇒ S_n = 11[4 + 147]
⇒ S_n = 11 × 151
⇒ S_n = 1661.

Hence, the sum of first 22 terms of the A.P. is 1661.

If first term is a and common difference is d of the A.P.

By formula,

n^th term = a_n = a + (n - 1)d

Given,

⇒ 5^th term = 4

⇒ a₅ = 4

⇒ a + (5 - 1)d = 4

⇒ a + 4d = 4 .........(1)

⇒ 9^th term = -12

⇒ a₉ = -12

⇒ a + (9 - 1)d = -12

⇒ a + 8d = -12 .........(2)

Subtracting equation (1) from (2), we get :

⇒ (a + 8d) - (a + 4d) = -12 - 4

⇒ a - a + 8d - 4d = -16

⇒ 4d = -16

⇒ d = $-\frac{16}{4}$

⇒ d = -4.

Substituting value of d in equation (1), we get :

⇒ a + 4d = 4

⇒ a + 4(-4) = 4

⇒ a - 16 = 4

⇒ a = 4 + 16 = 20.

(a) Hence, first term of A.P. = 20.

(b) Common difference of A.P. = -4.

(c) By formula,

Sum of n terms of A.P. = $\frac{n}{2}(a + l)$

Sum of first 20 terms of A.P. = $\frac{n}{2}(a + a_{20})$

$$\begin{aligned}= \frac{20}{2}[a + a + (20 - 1)d] \\[1em] = 10(2a + 19d) \\[1em] = 10 \times (2 \times 20 + 19 \times -4) \\[1em] = 10 \times (40 - 76) \\[1em] = 10 \times -36 \\[1em] = -360.\end{aligned}$$

Hence, sum of first 20 terms of A.P. = -360.

Given, a₂ = 14 and a₃ = 18.

common difference = d = any term - preceding term = a₃ - a₂ = 18 - 14 = 4.

By formula, a_n = a + (n - 1)d
⇒ a₂ = a + 4(2 - 1)
⇒ 14 = a + 4
⇒ a = 10.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

$$\begin{aligned}\Rightarrow S_{51} = \frac{51}{2}[2 \times 10 + (51 - 1)4] \\[1em] = \frac{51}{2}[20 + 50 \times 4] \\[1em] = \frac{51}{2}[20 + 200] \\[1em] = \frac{51}{2} \times 220 \\[1em] = 51 \times 110 \\[1em] = 5610.\end{aligned}$$

Hence, the sum of first 51 terms of the A.P. is 5610.

Given, a₄ = 22 and a₁₅ = 66.

By formula, a_n = a + (n - 1)d
⇒ a₄ = a + (4 - 1)d
⇒ 22 = a + 3d
⇒ a = 22 - 3d         (Eq 1)

⇒ a₁₅ = a + (15 - 1)d
⇒ 66 = a + 14d
Putting value of a from Eq 1 in above equation
⇒ 66 = 22 - 3d + 14d
⇒ 66 = 22 + 11d
⇒ 11d = 66 - 22
⇒ 11d = 44
⇒ d = 4.

Putting value of d in Eq 1,
⇒ a = 22 - 3(4)
⇒ a = 22 - 12
⇒ a = 10.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ S₈ = $\frac{8}{2}[2 \times 10 + (8 - 1)4]$
⇒ S₈ = 4[20 + 28]
⇒ S₈ = 4 × 48
⇒ S₈ = 192.

Hence, first term = a = 10, common difference = d = 4 and sum of first 8 terms = S₈ = 192.

Given, S₆ = 36 and S₁₆ = 256.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ S₆ = $\frac{6}{2}[2a + (6 - 1)d]$
⇒ 36 = 3[2a + 5d]
⇒ 2a + 5d = 12
⇒ 2a = 12 - 5d     (Eq 1)

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ S₁₆ = $\frac{16}{2}[2a + (16 - 1)d]$
⇒ 256 = 8[2a + 15d]
⇒ 2a + 15d = 32
⇒ 2a = 32 - 15d

Putting value of 2a from Eq 1 in above equation,

⇒ 12 - 5d = 32 - 15d
⇒ -5d + 15d = 32 - 12
⇒ 10d = 20
⇒ d = 2.

∴ From Eq 1,
⇒ 2a = 12 - 5d
⇒ 2a = 12 - 5(2)
⇒ 2a = 2
⇒ a = 1.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ S₁₀ = $\frac{10}{2}[2 \times 1 + (10 - 1)2]$
⇒ S₁₀ = 5[2 + 18]
⇒ S₁₀ = 5 × 20
⇒ S₁₀ = 100.

Hence, the sum of first 10 terms of the A.P. is 100.

a_n = 3 + 4n
a₁ = 3 + 4 × 1 = 3 + 4 = 7
a₂ = 3 + 4 × 2 = 3 + 8 = 11
a₃ = 3 + 4 × 3 = 3 + 12 = 15
a₄ = 3 + 4 × 4 = 3 + 16 = 19.

Since, a₄ - a₃ = a₃ - a₂ = a₂ - a₁ = 4, i.e any term - preceding term = fixed number = 4.

∴ a₁, a₂, a₃ ..... form an A.P.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

$$\begin{aligned}\Rightarrow S_{15} = \frac{15}{2}[2 \times 7 + (15 - 1)4] \\[1em] = \frac{15}{2}[14 + 14 \times 4] \\[1em] = \frac{15}{2}[14 + 56] \\[1em] = \frac{15}{2} \times 70 \\[1em] = 15 \times 35 \\[1em] = 525.\end{aligned}$$

Hence, the sum of first 15 terms of the A.P. is 525.

Given, S₆ = 42 and a₁₀ : a₃₀ = 1 : 3

$$\begin{aligned}\Rightarrow a_{10} : a_{30} = 1 : 3 \\[1em] \Rightarrow \frac{a_{10}}{a_{30}} = \frac{1}{3} \\[1em] \Rightarrow 3a_{10} = a_{30}\end{aligned}$$

By formula, a_n = a + (n - 1)d,

$$\begin{aligned}\Rightarrow 3[a + (10 - 1)d] = a + (30 - 1)d \\[1em] \Rightarrow 3(a + 9d) = a + 29d \\[1em] \Rightarrow 3a + 27d = a + 29d \\[1em] \Rightarrow 3a - a = 29d - 27d \\[1em] \Rightarrow 2a = 2d \\[1em] \Rightarrow a = d.\end{aligned}$$

S₆ = 42 or,

⇒ $\frac{6}{2}[2a + (6 - 1)d]$ = 42

Since, a = d
⇒ 3[2d + 5d] = 42
⇒ 3 × 7d = 42
⇒ 21d = 42
⇒ d = 2.

Since, a = d hence a = 2.

By formula, a_n = a + (n - 1)d
⇒ a₁₃ = 2 + (13 - 1)(2)
⇒ a₁₃ = 2 + 24
⇒ a₁₃ = 26.

Hence, the first term of the A.P. is 2 and the thirteenth term is 26.

Given, S_n = 6n - n².

So, sum of (n -1) terms will be

⇒ S_{n - 1} = 6(n - 1) - (n - 1)²
⇒ S_{n - 1} = 6n - 6 -(n² + 1 - 2n)
⇒ S_{n - 1} = 6n - 6 - n² - 1 + 2n
⇒ S_{n - 1} = 8n - n² - 7.

By formula, a_n = S_n - S_{n - 1}
⇒ a_n = 6n - n² - (8n - n² - 7)
⇒ a_n = 6n - 8n - n² + n² + 7
⇒ a_n = -2n + 7.

∴ a₂₅ = -2(25) + 7 = -50 + 7 = -43.

Hence, the 25th term of the A.P. is -43.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

We need to prove S₃₀ = 3(S₂₀ - S₁₀).

$$\begin{aligned}\Rightarrow \text{L.H.S.} = S_{30} = \frac{30}{2}[2a + (30 - 1)d] \\[1em] = 15[2a + 29d] \\[1em] = 30a + 435d \\[1em] \Rightarrow \text{R.H.S.} = 3(S_{20} - S_{10}) \\[1em] = 3\Big(\frac{20}{2}[2a + (20 - 1)d] - \frac{10}{2}[2a + (10 - 1)d]\Big) \\[1em] = 3\Big(10[2a + 19d] - 5[2a + 9d]\Big) \\[1em] = 3(20a + 190d - 10a - 45d) \\[1em] = 3(10a + 145d) \\[1em] = 30a + 435d.\end{aligned}$$

∴ L.H.S. = R.H.S. = 30a + 435d.

Hence, proved that S₃₀ = 3(S₂₀ - S₁₀).

We need to find 1 + 2 + 3 + 4 + ...... + 1000

The above series is an A.P. with first term a = 1, d = 1 and n = 1000.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ S₁₀₀₀ = $\frac{1000}{2}[2 \times 1 + (1000 - 1)1]$
⇒ S₁₀₀₀ = 500[2 + 999]
⇒ S₁₀₀₀ = 500 × 1001
⇒ S₁₀₀₀ = 500500.

Hence, the sum of first 1000 positive integers is 500500.

We need to find 8 + 16 + 24 + ..... + (15th term)

The above series forms an A.P. with a = 8 and d = 8.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ S₁₅ = $\frac{15}{2}[2 \times 8 + (15 - 1)8]$
⇒ 2S₁₅ = 15[16 + 14(8)]
⇒ 2S₁₅ = 15[128]
⇒ 2S₁₅ = 1920
⇒ S₁₅ = 960

Hence, the sum of first 15 terms is 960.

The sum of series of the two digit natural numbers that are divisible by 4 is $12 + 16 + 20 + .... + 96.$

a = 12, d = 16 - 12 = 4 and l = 96.

Let 96 be nth term then,

⇒ 96 = 12 + 4(n - 1)
⇒ 96 - 12 = 4n - 4
⇒ 84 = 4n - 4
⇒ 4n = 84 + 4
⇒ 4n = 88
⇒ n = 22.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ S₂₂ = $\frac{22}{2}[2 \times 12 + 4(22 - 1)]$
⇒ S₂₂ = 11[24 + 84]
⇒ S₂₂ = 11 × 108
⇒ S₂₂ = 1188.

Hence, the sum of all two digits natural numbers which are divisible by 4 is 1188.

The sum of natural numbers between 100 and 200 that are divisible by 4 is 104 + 108 + 112 + ..... + 196.

The above series is an A.P. with a = 104, d = 108 - 104 = 4 and l = 196.

Let 196 be nth term then,

⇒ 196 = 104 + 4(n - 1)
⇒ 196 - 104 = 4n - 4
⇒ 92 = 4n - 4
⇒ 4n = 92 + 4
⇒ 4n = 96
⇒ n = 24.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ S₂₄ = $\frac{24}{2}[2 \times 104 + 4(24 - 1)]$
⇒ S₂₄ = 12[208 + 92]
⇒ S₂₄ = 12 × 300
⇒ S₂₄ = 3600.

Hence, the sum of all two digits natural numbers between 100 and 200 which are divisible by 4 is 3600.

The sum of all multiples of 9 lying between 300 and 700 is 306 + 315 + ..... + 693.

The above series is an A.P. with a = 306, d = 9 and l = 693.

Let 693 be nth term of the series then,

⇒ 693 = 306 + 9(n - 1)
⇒ 693 - 306 = 9n - 9
⇒ 387 = 9n - 9
⇒ 9n = 387 + 9
⇒ 9n = 396
⇒ n = 44.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ S₄₄ = $\frac{44}{2}[2 \times 306 + 9(44 - 1)]$
⇒ S₄₄ = 22[612 + 387]
⇒ S₄₄ = 22 × 999
⇒ S₄₄ = 21978.

Hence, the sum of all multiples of 9 lying between 300 and 700 is 21978.

The sum of all natural numbers less than 100 which are divisible by 6 is given as 6 + 12 + 18 + .... + 96.

The above series is an A.P. with a = 6, d = 6 and l = 96.

Let 96 be nth term of the series then,

⇒ 96 = 6 + 6(n - 1)
⇒ 96 - 6 = 6n - 6
⇒ 90 = 6n - 6
⇒ 6n = 90 + 6
⇒ 6n = 96
⇒ n = 16.

By formula S_n = $\frac{n}{2}[2a + (n - 1)d]$

⇒ S₁₆ = $\frac{16}{2}[2 \times 6 + 6(16 - 1)]$
⇒ S₁₆ = 8[12 + 90]
⇒ S₁₆ = 8 × 102
⇒ S₁₆ = 816.

Hence, the sum of all natural numbers less than 100 which are divisible by 6 is 816.

Let common difference be d.

a = 3

Sum of first n terms of an A.P. = $\frac{\text{n}}{2}(\text{a } + \text{ l})$

Given,

The sum of the first 8 terms is twice the sum of the first 5 terms.

$$\begin{aligned}\therefore \frac{8}{2}(a + a_8) = 2 \times \frac{5}{2}(a + a_5) \\[1em] \Rightarrow 4[a + a + (8 - 1)d] = 5[a + a + (5 - 1)d] \\[1em] \Rightarrow 4[2a + 7d] = 5[2a + 4d] \\[1em] \Rightarrow 4[2 \times 3 + 7d] = 5[2 \times 3 + 4d] \\[1em] \Rightarrow 4[6 + 7d] = 5[6 + 4d] \\[1em] \Rightarrow 24 + 28d = 30 + 20d \\[1em] \Rightarrow 28d - 20d = 30 - 24 \\[1em] \Rightarrow 8d = 6 \\[1em] \Rightarrow d = \frac{6}{8} = \frac{3}{4}.\end{aligned}$$

Hence, common difference = $\frac{3}{4}$.

The given list of numbers $\frac{1}{6}, \frac{1}{3}, \frac{2}{3}, ...$ is a G.P. with first term a = $\frac{1}{6}$ and common ratio = r = 2.

By formula, a_n = ar^{n - 1}

∴ a₄ = $\frac{1}{6}(2)^3 = \frac{8}{6} = \frac{4}{3}.$

Hence, the next term of the G.P. is $\frac{4}{3}$.

The given list of numbers,

$\frac{3}{16}, -\frac{3}{8}, \frac{3}{4}, -\frac{3}{2}, ...$ is a G.P with first term a = $\frac{3}{16}$ and common ratio = r = -2.

By formula, a_n = ar^{n - 1}

∴ a₅ = $\frac{3}{16}(-2)^4 = \frac{3}{16} \times 16 = 3.$

Hence, the next term of the G.P. is 3.

The given list of numbers,

$\sqrt{3} + \frac{1}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + ....$ is a G.P. with first term a = $\sqrt{3}$ and common ratio = r = $\frac{1}{3}.$

By formula, a_n = ar^{n - 1}

∴ a₁₅ = $\sqrt{3}\Big(\frac{1}{3}\Big)^{14} = 3^{1/2}\Big(\frac{1}{3^{14}}\Big) = 3^{1/2}\times 3^{-14} = 3^{1/2 - 14} = 3^{-27/2} .$

Hence, the 15th term of the G.P. is $3^{-27/2}$.

The given list of numbers 5, 25, 125 .... is a G.P. with first term a = 5 and r = 5.

By formula, a_n = ar^{n - 1}

$$\begin{aligned}a_{10} = 5(5)^{10 - 1} = 5(5)^9 = 5^{10}. \\[1em] a_n = 5(5)^{n - 1} = 5 \times 5^n \times 5^{-1} = 5^n.\end{aligned}$$

Hence, the 10th term is 5¹⁰ and nth term is 5ⁿ.

The given list of numbers

$\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, ...$ is a G.P. with first term a = $\frac{3}{2}$ and common ratio = r = $\frac{1}{2}.$

By formula, a_n = ar^{n - 1}

$$\begin{aligned}a_6 = \frac{3}{2}\Big(\frac{1}{2}\Big)^{6 - 1} = \frac{3}{2}\Big(\frac{1}{2}\Big)^5 = \frac{3}{2} \times \frac{1}{32} = \frac{3}{64}. \\[1em] a_n = \frac{3}{2}\Big(\frac{1}{2}\Big)^{n - 1} = \frac{3}{2 \times 2^n \times 2^{-1}} = \frac{3}{2^n}.\end{aligned}$$

Hence, the 6th term is $\frac{3}{64} \text{ and nth term is } \frac{3}{2^n}.$

The given list of numbers 3, -6, 12, -24, .... , 12288 is a G.P. with last term = l = 12288 and the common ratio = r = -2.

By formula, nth term from end = $l\Big(\frac{1}{r}\Big)^{n - 1}$

$$\begin{aligned}\text{7th term from end } = 12288\Big(-\frac{1}{2}\Big)^{6 - 1} \\[1em] = 12288\Big(-\frac{1}{2}\Big)^5 \\[1em] = -\frac{12288}{32} \\[1em] = -384.\end{aligned}$$

Hence, the 6th term from end of the G.P. is -384.

(i) Given a = 2, r = $\sqrt{2}$.

Let nth term be 128.

By formula, a_n = ar^{n - 1}.

$$\begin{aligned}\Rightarrow 128 = 2(\sqrt{2})^{n - 1} \\[1em] \Rightarrow \frac{128}{2} = (2)^{(n - 1)/2} \\[1em] \Rightarrow 64 = (2)^{(n - 1)/2} \\[1em] \Rightarrow 2^6 = (2)^{(n - 1)/2} \\[1em] \Rightarrow \frac{n - 1}{2} = 6 \\[1em] \Rightarrow n - 1 = 12 \\[1em] \Rightarrow n = 13.\end{aligned}$$

Hence, 128 is 13th term of the G.P.

(ii) Given a = 1, r = $\frac{1}{3}$.

Let nth term be $\frac{1}{243}$.

By formula, a_n = ar^{n - 1}.

$$\begin{aligned}\Rightarrow \frac{1}{243} = 1\Big(\frac{1}{3}\Big)^{n - 1} \\[1em] \Rightarrow \frac{1}{3^5} = \frac{1}{3^{(n - 1)}} \\[1em] \Rightarrow n - 1 = 5 \\[1em] \Rightarrow n = 6.\end{aligned}$$

Hence, $\frac{1}{243}$ is 6th term of the G.P.

Given, a₈ = 192 and r = 2.

By formula, a_n = ar^{n - 1}.

⇒ a₈ = a(2)^{(8 - 1)}
⇒ 192 = a(2)⁷
⇒ a = $\frac{192}{2^7} = \frac{192}{128} = \frac{3}{2}.$

12th term of the G.P. is a₁₂,

$$\begin{aligned}\Rightarrow a_{12} = \frac{3}{2}(2)^{12 - 1} \\[1em] \Rightarrow a_{12} = \frac{3}{2} \times 2^{11} \\[1em] \Rightarrow a_{12} = 3 \times 2^{10} \\[1em] \Rightarrow a_{12} = 3 \times 1024 \\[1em] \Rightarrow a_{12} = 3072.\end{aligned}$$

Hence, the 12th term of the G.P. is 3072.

Given, a₃ = 24, a₆ = 192.

By formula, a_n = ar^{n - 1}.

⇒ a₃ = ar^{(3 - 1)}
⇒ ar² = 24.             (Eq 1)

⇒ a₆ = ar^{(6 - 1)}
⇒ ar⁵ = 192.            (Eq 2)

Dividing Eq 2 by Eq 1

$$\begin{aligned}\Rightarrow \frac{ar^5}{ar^2} = \frac{192}{24} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = (2)^3 \\[1em] \Rightarrow r = 2\end{aligned}$$

Putting value of r in Eq 1,

$$\begin{aligned}\Rightarrow a(2)^2 = 24 \\[1em] \Rightarrow 4a = 24 \\[1em] \Rightarrow a = 6 \\[1em] \text{10th term of G.P.} = a_{10} \\[1em] \Rightarrow a_{10} = 6(2)^{10 - 1} \\[1em] = 6(2)^9 = 6 \times 512 = 3072.\end{aligned}$$

Hence, the 10th term of the G.P. is 3072.

Let the number of terms be n.

Given, a = $\frac{3}{4}$, r = 2 and a_n = 384.

By formula, a_n = ar^{n - 1}.

$$\begin{aligned}\Rightarrow 384 = \frac{3}{4}(2)^{n - 1} \\[1em] \Rightarrow \frac{384 \times 4}{3} = (2)^{n - 1} \\[1em] \Rightarrow (2)^{n - 1} = 512 \\[1em] \Rightarrow (2)^{n - 1} = (2)^9 \\[1em] \Rightarrow n - 1 = 9 \\[1em] \Rightarrow n = 10.\end{aligned}$$

Hence, the number of terms in the G.P. are 10.

(i) Since, $-\frac{2}{7}, x, -\frac{7}{2}$ are three consecutive terms of a G.P.

So,

$$\begin{aligned}\Rightarrow \frac{x}{-\frac{2}{7}} = r = \frac{-\frac{7}{2}}{x} \\[1em] \Rightarrow \frac{x}{-\frac{2}{7}} = \frac{-\frac{7}{2}}{x} \\[1em] \Rightarrow x^2 = -\frac{7}{2} \times -\frac{2}{7} \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x^2 - 1 = 0 \\[1em] \Rightarrow (x - 1)(x + 1) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x + 1 = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = -1.\end{aligned}$$

Hence, the value of x = 1 or -1.

(ii) Since, x + 9, x - 6 and 4 are three consecutive terms of a G.P.

So,

$$\begin{aligned}\Rightarrow \frac{x - 6}{x + 9} = r = \frac{4}{x - 6} \\[1em] \Rightarrow \frac{x - 6}{x + 9} = \frac{4}{x - 6} \\[1em] \Rightarrow (x - 6)^2 = 4(x + 9) \\[1em] \Rightarrow x^2 + 36 - 12x = 4x + 36 \\[1em] \Rightarrow x^2 - 12x - 4x + 36 - 36 = 0 \\[1em] \Rightarrow x^2 - 16x = 0 \\[1em] \Rightarrow x(x - 16) = 0 \\[1em] \Rightarrow x = 0 \text{ or } x - 16 = 0 \\[1em] \Rightarrow x = 0 \text{ or } x = 16.\end{aligned}$$

Hence, the value of x = 0 or 16.

(iii) Since, x, x + 3 and x + 9 are first three terms of a G.P.

So,

$$\begin{aligned}\Rightarrow \frac{x + 3}{x} = r = \frac{x + 9}{x + 3} \\[1em] \Rightarrow \frac{x + 3}{x} = \frac{x + 9}{x + 3} \\[1em] \Rightarrow (x + 3)^2 = x(x + 9) \\[1em] \Rightarrow x^2 + 9 + 6x = x^2 + 9x \\[1em] \Rightarrow x^2 - x^2 + 9 + 6x - 9x = 0 \\[1em] \Rightarrow 9 - 3x = 0 \\[1em] \Rightarrow 3x = 9 \\[1em] \Rightarrow x = 3. \\[1em]\end{aligned}$$

Hence, the value of x = 3.

Given, a₄ = x, a₇ = y and a₁₀ = z.

By formula, a_n = ar^{n - 1}.

⇒ x = a₄ = a(r)³

⇒ y = a₇ = a(r)⁶

⇒ z = a₁₀ = a(r)⁹

From above equations,

$$\begin{aligned}\Rightarrow \frac{y}{x} = \frac{ar^6}{ar^3} = r^3. \\[1em] \Rightarrow \frac{z}{y} = \frac{ar^9}{ar^6} = r^3.\end{aligned}$$

The above equations prove that the common ratio between x, y and z is r³.

Hence, x, y and z are in G.P. with common ratio = r³.

Let the first term of the G.P. be a and common ratio = r.

Given,

⇒ a₅ = ar⁴ = p

⇒ a₈ = ar⁷ = q

⇒ a₁₁ = ar¹⁰ = s

We need to prove q² = ps.

L.H.S. = q²

⇒ q² = q × q = ar⁷ × ar⁷ = a²r¹⁴.

R.H.S. = ps

⇒ ps = p × s = ar⁴ × ar¹⁰ = a²r¹⁴.

∴ L.H.S. = R.H.S. = a²r¹⁴.

Hence, proved that q² = ps.

Since a, a² + 2 and a³ + 10 are in G.P.

$$\begin{aligned}\therefore \frac{a^2 + 2}{a} = r = \frac{a^3 + 10}{a^2 + 2} \\[1em] \Rightarrow \frac{a^2 + 2}{a} = \frac{a^3 + 10}{a^2 + 2} \\[1em] \Rightarrow (a^2 + 2)^2 = a(a^3 + 10) \\[1em] \Rightarrow a^4 + 4 + 4a^2 = a^4 + 10a \\[1em] \Rightarrow a^4 - a^4 + 4a^2 - 10a + 4 = 0 \\[1em] \Rightarrow 4a^2 - 10a + 4 = 0 \\[1em] \Rightarrow 4a^2 - 8a - 2a + 4 = 0 \\[1em] \Rightarrow 4a(a - 2) - 2(a - 2) = 0 \\[1em] \Rightarrow (4a - 2)(a - 2) = 0 \\[1em] \Rightarrow 4a - 2 = 0 \text{ or } a - 2 = 0 \\[1em] \Rightarrow a = \frac{2}{4} \text{ or } a = 2. \\[1em] \Rightarrow a = \frac{1}{2} \text{ or } a = 2.\end{aligned}$$

Hence, the required value(s) of a are $\frac{1}{2}$ and 2.

Given, a₄ = 54 and a₇ = 1458.

By formula, a_n = ar^{n - 1}.

⇒ a₄ = a(r)^{4 - 1}
⇒ 54 = ar³     (Eq 1)

⇒ a₇ = a(r)^{7 - 1}
⇒ 1458 = a(r)⁶     (Eq 2)

Dividing Eq 2 by Eq 1,

$$\begin{aligned}\Rightarrow \frac{ar^6}{ar^3} = \frac{1458}{54} \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r = \sqrt[3]{27} \\[1em] \Rightarrow r = 3\end{aligned}$$

Putting value of r in Eq 1,

$$\begin{aligned}\Rightarrow a(3)^3 = 54 \\[1em] \Rightarrow 27a = 54 \\[1em] \Rightarrow a = 2.\end{aligned}$$

a₂ = ar = 2 × 3 = 6
a₃ = ar² = 2(3)² = 2 × 9 = 18
a₄ = ar³ = 2(3)³ = 2 × 27 = 54.

Hence, the required G.P. is 2, 6, 18, 54, ...

Given, S₃ = $\frac{39}{10}$ and a₁ × a₂ × a₃ = 1.

Let the three numbers that are in G.P. be $\frac{a}{r}, a, ar.$

$$\begin{aligned}\therefore \frac{a}{r} \times a \times ar = 1 \\[1em] \Rightarrow a^3 = 1 \\[1em] \Rightarrow a = 1. \\[1em] S_3 = \frac{39}{10} \\[1em] \Rightarrow \frac{a}{r} + a + ar = \frac{39}{10} \\[1em] \Rightarrow a\Big(\frac{1}{r} + 1 + r\Big) = \frac{39}{10} \\[1em] \Rightarrow 1\Big(\frac{1 + r + r^2}{r}\Big) = \frac{39}{10} \\[1em] \Rightarrow 10(1 + r + r^2) = 39r \\[1em] \Rightarrow 10 + 10r + 10r^2 = 39r \\[1em] \Rightarrow 10r^2 + 10r - 39r + 10 = 0 \\[1em] \Rightarrow 10r^2 - 29r + 10 = 0 \\[1em] \Rightarrow 10r^2 - 25r - 4r + 10 = 0 \\[1em] \Rightarrow 5r(2r - 5) - 2(2r - 5) = 0 \\[1em] \Rightarrow (5r - 2)(2r - 5) = 0 \\[1em] \Rightarrow 5r - 2 = 0 \text{ or } 2r - 5 = 0 \\[1em] \Rightarrow r = \frac{2}{5} \text{ or } r = \frac{5}{2} \\[1em] \text{First taking r } = \frac{2}{5} \\[1em] \therefore \text{Terms are : } \frac{a}{r} = \frac{1}{\frac{2}{5}} = \frac{5}{2}, a = 1, ar = 1 \times \frac{2}{5} = \frac{2}{5}. \\[1em] \text{Now, taking r } = \frac{5}{2} \\[1em] \therefore \text{Terms are : } \frac{a}{r} = \frac{1}{\frac{5}{2}} = \frac{2}{5}, a = 1, ar = 1 \times \frac{5}{2} = \frac{5}{2}. \\[1em]\end{aligned}$$

Hence, common ratio is $\frac{2}{5}$ or $\frac{5}{2}$ and terms are $\frac{5}{2}, 1, \frac{2}{5}$ or $\frac{2}{5}, 1, \frac{5}{2}.$

Let three numbers that are in A.P. be a - d, a, a + d.

Given, sum of three numbers = 15.

⇒ a - d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5.

By adding 1, 4 and 19 in the terms become,

⇒ a - d + 1, a + 4 and a + d + 19
⇒ 5 - d + 1, 5 + 4 and 5 + d + 19
⇒ 6 - d, 9 and d + 24.

According to question these terms become in G.P.,

$$\begin{aligned}\therefore \frac{9}{6 - d} = \frac{d + 24}{9} \\[1em] \Rightarrow (d + 24)(6 - d) = 81 \\[1em] \Rightarrow 6d - d^2 + 144 - 24d = 81 \\[1em] \Rightarrow 6d - d^2 - 24d + 144 - 81 = 0 \\[1em] \Rightarrow -18d - d^2 + 63 = 0 \\[1em] \Rightarrow d^2 + 18d - 63 = 0 \\[1em] \Rightarrow d^2 + 21d - 3d - 63 = 0 \\[1em] \Rightarrow d(d + 21) - 3(d + 21) = 0 \\[1em] \Rightarrow (d - 3)(d + 21) = 0 \\[0.5em] \Rightarrow d - 3 = 0 \text{ or } d + 21 = 0 \\[1em] \Rightarrow d = 3 \text{ or } d = -21.\end{aligned}$$

Taking d = 3,

∴ a - d = 5 - 3 = 2, a = 5, a + d = 5 + 3 = 8.

Taking d = - 21,

∴ a - d = 5 -(-) 21 = 26, a = 5, a + d = 5 +(-) 21 = - 16.

Hence, the required numbers are 2, 5 and 8 or 26, 5 and -16.

(i) Sum = 2 + 6 + 18 + ..... + (20th term)

The above list of numbers is a G.P. with first term a = 2 and common ratio = r = $\frac{6}{2} = 3.$

By formula,

$$\begin{aligned}S_n = \frac{a(r^n - 1)}{r - 1} \\[1em] \therefore S_{20} = \frac{2((3)^{20} - 1)}{3 - 1} \\[1em] = \frac{2(3^{20} - 1)}{2} \\[1em] = 3^{20} - 1.\end{aligned}$$

Hence, the sum of the series is 3²⁰ - 1.

(ii) Sum = 1 + $\sqrt{3}$ + 3 + .... + (10th term)

The above list of numbers is a G.P. with first term a = 1 and common ratio = r = $\sqrt{3}.$

By formula,

$$\begin{aligned}S_n = \frac{a(r^n - 1)}{r - 1} \\[1em] \therefore S_{10} = \frac{1((\sqrt{3})^{10} - 1)}{\sqrt{3} - 1} \\[1em] = \frac{((3)^{10/2} - 1)}{\sqrt{3} - 1} \\[1em] = \frac{3^5 - 1}{\sqrt{3} - 1}\end{aligned}$$

Multiplying numerator and denominator by $\sqrt{3}$ + 1,

$$\begin{aligned}= \frac{3^5 - 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] = \frac{242(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\[1em] = \frac{242(\sqrt{3} + 1)}{3 + \sqrt{3} - \sqrt{3} - 1} \\[1em] = \frac{242(\sqrt{3} + 1)}{2} \\[1em] = 121(\sqrt{3} + 1).\end{aligned}$$

Hence, the sum of the series is $121(\sqrt{3} + 1)$.

(iii) Sum = 1, $-\frac{2}{3}, \frac{4}{9},$ ....(6th term)

The above list of numbers is a G.P. with first term a = 1 and common ratio = r = $-\frac{2}{3}.$

By formula,

$$\begin{aligned}S_n = \frac{a(r^n - 1)}{r - 1} \\[1em] \therefore S_{6} = \frac{1\Big[\Big(-\frac{2}{3}\Big)^6 - 1\Big]}{-\frac{2}{3} - 1} \\[1em] = \frac{\frac{2^6}{3^6} - 1}{\frac{-2 -3}{3}} \\[1em] = \frac{\frac{2^6}{3^6} - 1}{-\frac{5}{3}} \\[1em] = \frac{3(2^6 - 3^6)}{3^6 \times -5} \\[1em] = \frac{64 - 729}{3^5 \times -5} \\[1em] = \frac{-665}{-1215}\end{aligned}$$

On dividing numerator and denominator by -5 we get,

$= \frac{133}{243}.$

Hence, the sum of the series is $\frac{133}{243}$.

(iv) Sum upto 5 terms = 1 + $\frac{2}{3} + \frac{4}{9} + ....$ + 5th term

The above list of numbers is a G.P. with first term a = 1 and common ratio = r = $\frac{2}{3}.$

By formula,

$$\begin{aligned}S_n = \frac{a(r^n - 1)}{r - 1} \\[1em] \therefore S_5 = \frac{1\Big[\Big(\frac{2}{3}\Big)^5 - 1\Big]}{\frac{2}{3} - 1} \\[1em] = \frac{\frac{2^5}{3^5} - 1}{\frac{2 -3}{3}} \\[1em] = \frac{\frac{2^5}{3^5} - 1}{-\frac{1}{3}} \\[1em] = \frac{3(2^5 - 3^5)}{3^5 \times -1} \\[1em] = \frac{32 - 243}{3^4 \times -1} \\[1em] = \frac{-211}{-81}\end{aligned}$$

On dividing numerator and denominator by -1 we get,

$= \frac{211}{81}$

Sum of n terms = $S_n = \frac{1\Big[\Big(\frac{2}{3}\Big)^n - 1\Big]}{\frac{2}{3} - 1}$

$$\begin{aligned}= \frac{\Big(\frac{2}{3}\Big)^n - 1}{\frac{2 - 3}{3}} \\[1em] = \frac{3\Big[\Big(\frac{2}{3}\Big)^n - 1\Big]}{-1} \\[1em] = -3\Big[\Big(\frac{2}{3}\Big)^n - 1\Big] \\[1em] = 3\Big[1 - \Big(\frac{2}{3}\Big)^n\Big].\end{aligned}$$

Hence, the sum of the series upto 5 terms is $\frac{211}{81}$ and upto n terms is $3\Big[1 - \Big(\frac{2}{3}\Big)^n\Big]$.

The above series is in G.P. with a = 81, r = $-\frac{1}{3} \text{ and l } = -\frac{1}{27}.$

$$\begin{aligned}\text{Sum } = \frac{a - lr}{1 - r} \\[1em] \therefore \text{Sum } = \frac{81 - \Big(-\frac{1}{27}\Big)(-\frac{1}{3})}{1 - \Big(-\frac{1}{3}\Big)} \\[1em] = \frac{81 - \frac{1}{81}}{1 + \frac{1}{3}} \\[1em] = \frac{\frac{6561 - 1}{81}}{\frac{3 + 1}{3}} \\[1em] = \frac{\frac{6560}{81}}{\frac{4}{3}} \\[1em] = \frac{6560 \times 3}{81 \times 4} \\[1em] = \frac{19680}{324}\end{aligned}$$

Dividing numerator and denominator by 12 we get,

$= \frac{1640}{27}.$

Hence, the sum of the series 81 - 27 + 9 - ..... - $\frac{1}{27} \text{ is } \frac{1640}{27}.$

Given, a_n = 128, r = 2 and S_n = 255.

By formula a_n = ar^{n - 1}
∴ 128 = a(2)^{n - 1}

$$\begin{aligned}\Rightarrow a\frac{2^n}{2} = 128 \\[1em] \Rightarrow 2^n = \frac{256}{a}. \text{ (Eq 1) }\end{aligned}$$

By formula,

$$\begin{aligned}S_n = \frac{a(r^n - 1)}{r - 1} \\[1em] \therefore 255 = \frac{a(2^n - 1)}{2 - 1}\end{aligned}$$

Putting value of 2ⁿ from Eq 1 in above Equation,

$$\begin{aligned}\Rightarrow 255 = a\Big(\frac{256}{a} - 1\Big) \\[1em] \Rightarrow 255 = 256 - a \\[1em] \Rightarrow a = 256 - 255 = 1.\end{aligned}$$

Hence, the first term of the G.P. is 1.

The given series is a G.P. with a = 3 and r = 3.

Let n terms be required to give the sum of 120.

By formula,

$$\begin{aligned}S_n = \frac{a(r^n - 1)}{r - 1} \\[1em] \therefore 120 = \frac{3(3^n - 1)}{3 - 1} \\[1em] \Rightarrow 120 = \frac{3(3^n - 1)}{2} \\[1em] \Rightarrow 240 = 3(3^n - 1) \\[1em] \Rightarrow 80 = 3^n - 1 \\[1em] \Rightarrow 81 = 3^n \\[1em] \Rightarrow 3^4 = 3^n \\[1em] \Rightarrow n = 4.\end{aligned}$$

Hence, the required number of terms of the G.P. are 4.

The above series is a G.P. with a = 1 and r = 4.

Let n terms be required to give the sum of 341.

By formula,

$$\begin{aligned}S_n = \frac{a(r^n - 1)}{r - 1} \\[1em] \therefore 341 = \frac{1(4^n - 1)}{4 - 1} \\[1em] \Rightarrow 341 = \frac{(4^n - 1)}{3} \\[1em] \Rightarrow 341 \times 3 = (4^n - 1) \\[1em] \Rightarrow 1023 = 4^n - 1 \\[1em] \Rightarrow 4^n = 1024 \\[1em] \Rightarrow 4^n = 4^5 \\[1em] \therefore n = 5.\end{aligned}$$

Hence, the required number of terms of the G.P. are 5.

The above series is a G.P. with a = $\frac{2}{9} \text{ and } r = \frac{-\frac{1}{3}}{\frac{2}{9}} = -\frac{1}{3} \times \frac{9}{2} = -\frac{3}{2}$.

Let n terms be required to give the sum of $\frac{55}{72}$.

By formula,

$$\begin{aligned}S_n = \frac{a(r^n - 1)}{r - 1} \\[1em] \therefore \frac{55}{72} = \frac{\frac{2}{9}\Big[\Big(-\frac{3}{2}\Big)^n - 1\Big]}{-\frac{3}{2} - 1} \\[1em] \Rightarrow \frac{55}{72} = \frac{2\Big[\Big(-\frac{3}{2}\Big)^n - 1\Big]}{9 \times \Big(\frac{-3 - 2}{2}\Big)} \\[1em] \Rightarrow \frac{55}{72} = \frac{4\Big[\Big(-\frac{3}{2}\Big)^n - 1\Big]}{9 \times (-5)} \\[1em] \Rightarrow \frac{55}{72} = \frac{4\Big[\Big(-\frac{3}{2}\Big)^n - 1\Big]}{-45} \\[1em] \Rightarrow \frac{55 \times -45}{72 \times 4} = \Big(-\frac{3}{2}\Big)^n - 1 \\[1em] \Rightarrow -\frac{2475}{288} = \Big(-\frac{3}{2}\Big)^n - 1 \\[1em] \Rightarrow -\frac{2475}{288} + 1 = \Big(-\frac{3}{2}\Big)^n \\[1em] \Rightarrow \frac{-2475 + 288}{288} = \Big(-\frac{3}{2}\Big)^n \\[1em] \Rightarrow \Big(-\frac{3}{2}\Big)^n = -\frac{2187}{288} \\[1em]\end{aligned}$$

Dividing the numerator and denominator by 9

$$\begin{aligned}\Rightarrow \Big(-\frac{3}{2}\Big)^n = -\frac{243}{32} \\[1em] \Rightarrow \Big(-\frac{3}{2}\Big)^n = \Big(-\frac{3}{2}\Big)^5 \\[1em] \therefore n = 5.\end{aligned}$$

Hence, the required number of terms of the G.P. are 5.

Given,

$a_2 = -\frac{1}{2}$ and $a_5 = \frac{1}{16}$

By formula, $a_n = ar^{n - 1}$ we get,

$\Rightarrow a_2 = ar \text{ and } a_5 = ar^4$

Dividing $a_5$ by $a_2$,

$$\begin{aligned}\Rightarrow \frac{ar^4}{ar} = \frac{\frac{1}{16}}{-\frac{1}{2}} \\[1em] \Rightarrow r^3 = -\frac{1}{8} \\[1em] \Rightarrow r^3 = \Big(-\frac{1}{2}\Big)^3 \\[1em] \therefore r = -\frac{1}{2}.\end{aligned}$$

Since, $a_2 = ar$ or,

$$\begin{aligned}\Rightarrow -\frac{1}{2} = a\Big(-\frac{1}{2}\Big) \\[1em] \Rightarrow a = 1.\end{aligned}$$

By formula,

$$\begin{aligned}S_n = \frac{a(r^n - 1)}{r - 1} \\[1em] \therefore S_8 = \frac{1\Big[\Big(-\frac{1}{2}\Big)^8 - 1\Big]}{-\frac{1}{2} - 1} \\[1em] = \frac{\frac{1}{256} - 1}{\frac{-1 - 2}{2}} \\[1em] = \frac{\frac{1 - 256}{256}}{\frac{-3}{2}} \\[1em] = \frac{-255 \times 2}{256 \times (-3)} \\[1em] = \frac{-510}{-768}\end{aligned}$$

Dividing numerator and denominator by 6,

$= \frac{85}{128}.$

Hence, the sum of the series upto 8 terms is $\frac{85}{128}$.

Given, a = 27 and $a_8 = \frac{1}{81}.$

By formula,

$$\begin{aligned}a_n = ar^{n - 1} \\[1em] \therefore a_8 = ar^7 \\[1em] \Rightarrow \frac{1}{81} = 27r^7 \\[1em] \Rightarrow \frac{1}{81 \times 27} = r^7 \\[1em] \Rightarrow \frac{1}{3^4 \times 3^3} = r^7 \\[1em] \Rightarrow r^7 = \frac{1}{3^7} \\[1em] \Rightarrow r = \frac{1}{3}.\end{aligned}$$

By formula,

$$\begin{aligned}S_n = \frac{a(r^n - 1)}{r - 1} \\[1em] \therefore S_{10} = \frac{27\Big[\Big(\frac{1}{3}\Big)^{10} - 1\Big]}{\frac{1}{3} - 1} \\[1em] = \frac{27(\frac{1}{3^{10}} - 1)}{\frac{1 - 3}{3}} \\[1em] = \frac{27 \frac{(1 - 3^{10})}{3^{10}}}{-\frac{2}{3}} \\[1em] = \frac{3^3 \frac{(1 - 3^{10})}{3^{10}}}{-2} \times 3 \\[1em] = \frac{3^4(1 - 3^{10})}{-2 \times 3^{10}} \\[1em] = \frac{3^4}{2} \times \frac{-(1 - 3^{10})}{3^{10}} \\[1em] = \frac{81}{2} \Big(\frac{3^{10} - 1}{3^{10}}\Big) \\[1em] = \frac{81}{2} \Big(1 - \frac{1}{3^{10}}\Big).\end{aligned}$$

Hence, the sum of first 10 terms of G.P. is $\frac{81}{2} \Big(1 - \frac{1}{3^{10}}\Big).$

Given, r = 3, l = 486 and S_n = 728.

Let the last term be nth term.

Hence, l = a_n = 486.

Using formula,

$$\begin{aligned}a_n = ar^{n - 1} \\[1em] \Rightarrow 486 = a(3)^{n - 1} \\[1em] \Rightarrow a\frac{3^n}{3} = 486 \\[1em] \Rightarrow 3^na = 486 \times 3 \\[1em] \Rightarrow 3^na = 1458 \\[1em] \Rightarrow a = \frac{1458}{3^n}. \text{ (Eq 1)}\end{aligned}$$

Given, $S_n = 728$,

Putting value of $a = \frac{1458}{3^n}$ in formula $S_n = \frac{a(r^n - 1)}{r - 1}$ we get,

$$\begin{aligned}\Rightarrow S_n = \frac{\frac{1458}{3^n}(3^n - 1)}{3 - 1} \\[1em] \Rightarrow 728 = \frac{1458 - \frac{1458}{3^n}}{2} \\[1em] \Rightarrow 1458 - \frac{1458}{3^n} = 728 \times 2 \\[1em] \Rightarrow \frac{1458}{3^n} = 1458 - 1456 \\[1em] \Rightarrow \frac{1458}{3^n} = 2 \\[1em] \Rightarrow 3^n = \frac{1458}{2} \\[1em] \Rightarrow 3^n = 729\end{aligned}$$

Putting the above value in Eq 1,

$$\begin{aligned}\Rightarrow a = \frac{1458}{729} \\[1em] \therefore a = 2.\end{aligned}$$

Hence, the first term of the G.P. is 2.

Let nth term be the last term of the G.P.

Given, a = 7, l = a_n = 448 and S_n = 889.

Using formula,

$$\begin{aligned}a_n = ar^{n - 1} \\[1em] \Rightarrow 448 = 7(r)^{n - 1} \\[1em] \Rightarrow r^{n - 1} = 64 \\[1em] \Rightarrow \frac{r^n}{r} = 64 \\[1em] \Rightarrow r^n = 64r. \text{ (Eq 1)} \\[1em]\end{aligned}$$

Given, S_n = 889

Putting value of rⁿ from Eq 1 in the formula $S_n = \frac{a(r^n - 1)}{r - 1}$ we get,

$$\begin{aligned}\Rightarrow 889 = \frac{7(64r - 1)}{r - 1} \\[1em] \Rightarrow 889(r - 1) = 448r - 7 \\[1em] \Rightarrow 889r - 889 = 448r - 7 \\[1em] \Rightarrow 889r - 448r = 889 - 7 \\[1em] \Rightarrow 441r = 882 \\[1em] \therefore r = 2.\end{aligned}$$

Hence, the common ratio of the G.P. is 2.

Given, r = 3 and S₇ = 2186.

Using formula $S_n = \frac{a(r^n - 1)}{r - 1}$ we get,

$$\begin{aligned}\Rightarrow S_7 = \frac{a(3^7 - 1)}{3 - 1} \\[1em] \Rightarrow 2186 = \frac{a(2187 - 1)}{2} \\[1em] \Rightarrow \frac{2186a}{2} = 2186 \\[1em] \Rightarrow a = 2. \\[1em]\end{aligned}$$

∴ Third term = a₃ = ar² = 2(3)² = 18.

Hence, the third term of the G.P. is 18.

Given, a = 5 and S₃ = $\frac{31}{5}$.

Using formula $S_n = \frac{a(r^n - 1)}{r - 1}$ we get,

$$\begin{aligned}\Rightarrow S_3 = \frac{5(r^3 - 1)}{r - 1} \\[1em] \Rightarrow \frac{31}{5} = \frac{5(r^3 - 1)}{r - 1} \\[1em] \Rightarrow 31(r - 1) = 25(r^3 - 1) \\[1em] \Rightarrow 25(r - 1)(r^2 + r + 1) = 31(r - 1) \\[1em]\end{aligned}$$

Dividing by (r - 1) on both sides we get,

$$\begin{aligned}\Rightarrow 25(r^2 + r + 1) = 31 \\[1em] \Rightarrow 25r^2 + 25r + 25 = 31 \\[1em] \Rightarrow 25r^2 + 25r + 25 - 31 = 0 \\[1em] \Rightarrow 25r^2 + 25r - 6 = 0 \\[1em] \Rightarrow 25r^2 + 30r - 5r - 6 = 0 \\[1em] \Rightarrow 5r(5r + 6) - 1(5r + 6) = 0 \\[1em] \Rightarrow (5r - 1)(5r + 6) = 0 \\[1em] \Rightarrow 5r - 1 = 0 \text{ or } 5r + 6 = 0 \\[1em] \Rightarrow 5r = 1 \text{ or } 5r = -6 \\[1em] \Rightarrow r = \frac{1}{5} \text{ or } r = -\frac{6}{5}.\end{aligned}$$

Hence, the common ratio of the G.P. is $\frac{1}{5} \text{ or } -\frac{6}{5}.$

Let first term of G.P. be a and common ratio be r.

Given,

First term (a) = 24

Fifth term (ar⁴) = 8

$$\begin{aligned}\Rightarrow \frac{ar^4}{a} = \frac{8}{24} \\[1em] \Rightarrow r^4 = \frac{1}{3} \Rightarrow (r^4)^2 = \Big(\frac{1}{3}\Big)^2 \\[1em] \Rightarrow r^8 = \frac{1}{9}.\end{aligned}$$

By formula,

Ninth term of G.P. (a₉) = ar⁸

= $24 \times \frac{1}{9} = \frac{8}{3}$.

Hence, ninth term of G.P. = $\frac{8}{3}$.

The above series is an A.P. with first term = a = 5 and common difference = d = 8 - 5 = 3.

We know that

    a_n = a + (n - 1)d
∴ a₁₀ = 5 + (10 - 1) × 3 = 5 + 9 × 3 = 5 + 27 = 32.

Hence, Option 1 is the correct option.

The above series is an A.P. with first term = a = 10 and common difference = d = 7 - 10 = -3.

We know that

    a_n = a + (n - 1)d
∴ a₃₀ = 10 + (30 - 1) × (-3) = 10 + 29 × (-3) = 10 - 87 = -77.

Hence, Option 3 is the correct option.

The above series is an A.P. with first term = a = -3 and common difference = $-\frac{1}{2} - (-3) = -\frac{1}{2} + 3 = -\frac{-1 + 6}{2} = \frac{5}{2}$.

We know that

a_n = a + (n - 1)d

$$\begin{aligned}\therefore a_{11} = -3 + (11 - 1) \times \frac{5}{2} \\[1em] = -3 + 10 \times \frac{5}{2} \\[1em] = -3 + 25 \\[1em] = 22.\end{aligned}$$

Hence, Option 2 is the correct option.

Here, common difference = d = 10 - 7 = 3 and last term = l = 130.

We know that the nth term from the end is given by the formula:

    nth term from end = l - (n - 1)d
∴ 11th term from the end = 130 - (15 - 1) × 3 = 130 - 14 × 3 = 130 - 42 = 88.

Hence, Option 3 is the correct option.

Given d = 5,

We know that

    a_n = a + (n - 1)d
∴ a₁₈ = a + (18 - 1) × 5 = a + 17 × 5 = a + 85.
    a₁₃ = a + (13 - 1) × 5 = a + 12 × 5 = a + 60.

∴ a₁₈ - a₁₃ = a + 85 - (a + 60) = a - a + 85 - 60 = 25.

Hence, Option 3 is the correct option.

Given, a₁₈ - a₁₄ = 32.

We know that

    a_n = a + (n - 1)d
∴ a₁₈ = a + (18 - 1) × d = a + 17d.
    a₁₄ = a + (14 - 1) × d = a + 13d.

∴ a₁₈ - a₁₄ = a + 17d - (a + 13d) = a - a + 17d - 13d = 4d.

⇒ 4d = 32
⇒ d = 8.

Hence, Option 1 is the correct option.

We know that

    a_n = a + (n - 1)d
∴ 4 = a + (7 - 1) × (-4)
⇒ 4 = a + 6 × (-4)
⇒ 4 = a - 24
⇒ a = 24 + 4
⇒ a = 28.

Hence, Option 4 is the correct option.

We know that

    a_n = a + (n - 1)d
∴ a_n = 3.5 + (101 - 1) × 0
⇒ a_n = 3.5 + 100 × 0
⇒ a_n = 3.5

Hence, Option 2 is the correct option.

Here, a = 21, d = 42 - 21 = 21.

Let nth term be 210, so a_n = 210.

We know that

    a_n = a + (n - 1)d
∴ 210 = 21 + (n - 1) × 21
⇒ 210 = 21 + 21n - 21
⇒ 21n = 210
⇒ n = 10

Hence, Option 2 is the correct option.

Here, a = 5, d = 3 - 5 = -2.

Let nth term be -41, so a_n = -41.

We know that

    a_n = a + (n - 1)d
∴ -41 = 5 + (n - 1) × (-2)
⇒ -41 = 5 - 2n + 2
⇒ 7 - 2n = -41
⇒ 2n = 7 + 41
⇒ 2n = 48
⇒ n = 24.

Hence, Option 3 is the correct option.

We know that in an A.P.,

Common difference = d = any term - preceding term

∴ 2k + 3 - (k + 1) = k + 1 - (k - 1)
⇒ 2k - k + 3 - 1 = k - k + 1 - (-1)
⇒ k + 2 = 2
⇒ k = 0.

Hence, Option 2 is the correct option.

Given a = -3 and a₂ = 4

We know that

    a_n = a + (n - 1)d
∴ a₂ = -3 + (2 - 1) × d
⇒ 4 = -3 + d
⇒ d = 4 + 3
⇒ d = 7.

    a₂₁ = -3 + (21 - 1) × 7
⇒ a₂₁ = -3 + 20 × 7
⇒ a₂₁ = -3 + 140
⇒ a₂₁ = 137.

Hence, Option 2 is the correct option.

Given, a = -5 and d = 2.

We know that,

$$\begin{aligned}S_n = \frac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S_6 = \frac{6}{2}[2 \times -5 + (6 - 1) \times 2] \\[1em] = 3[-10 + 5 \times 2] \\[1em] = 3[-10 + 10] \\[1em] = 3 \times 0 \\[1em] = 0.\end{aligned}$$

Hence, Option 1 is the correct option.

The above series is an A.P. with first term = $-\frac{2}{3}$ and common difference = d = $-\frac{2}{3} - \Big(-\frac{2}{3}\Big) = 0$

We know that,

$$\begin{aligned}S_n = \frac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S_{25} = \frac{25}{2}\Big[2 \times -\frac{2}{3} + (25 - 1) \times 0\Big] \\[1em] = \frac{25}{2}\Big[-\frac{4}{3} + 24 \times 0\Big] \\[1em] = \frac{25}{2} \times -\frac{4}{3} \\[1em] = -\frac{50}{3}.\end{aligned}$$

Hence, Option 3 is the correct option.

Given a = 1, a_n = 20 and S_n = 399.

We know that

    a_n = a + (n - 1)d
∴ a_n = 1 + (n - 1) × d
⇒ 20 = 1 + (n - 1)d
⇒ (n - 1)d = 20 - 1
⇒ (n - 1)d = 19
⇒ d = $\frac{19}{n - 1}$.

The formula for sum of A.P. is given by,

$$\begin{aligned}S_n = \frac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S_n = \frac{n}{2}[2 \times 1 + (n - 1) \times d] \\[1em] \Rightarrow 399 = \frac{n}{2}[2 + (n - 1) \times \frac{19}{n - 1}] \\[1em] \Rightarrow 399 = \frac{n}{2}[2 + 19] \\[1em] \Rightarrow \frac{n}{2} \times 21 = 399 \\[1em] \Rightarrow n = \frac{399 \times 2}{21} \\[1em] \Rightarrow n = \frac{798}{21} \\[1em] \Rightarrow n = 38.\end{aligned}$$

Hence, Option 3 is the correct option.

We know that

    l = a + (n - 1)d
∴ 21 = -5 + (n - 1) × d
⇒ (n - 1)d = 21 + 5
⇒ (n - 1)d = 26
⇒ d = $\frac{26}{n - 1}$.

The formula for sum of A.P. is given by,

$$\begin{aligned}S_n = \frac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S_n = \frac{n}{2}[2 \times (-5) + (n - 1) \times \frac{26}{n - 1}] \\[1em] \Rightarrow 200 = \frac{n}{2}[-10 + (n - 1) \times \frac{26}{n - 1}] \\[1em] \Rightarrow 200 = \frac{n}{2}[-10 + 26] \\[1em] \Rightarrow \frac{n}{2} \times 16 = 200 \\[1em] \Rightarrow 8n = 200 \\[1em] \Rightarrow n = \frac{200}{8} \\[1em] \Rightarrow n = 25.\end{aligned}$$

Hence, Option 4 is the correct option.

First five multiples of 3 are : 3, 6, 9, 12, 15.

The above series is an A.P. with first term = a = 3 and common difference = d = 3.

The formula for sum of A.P. is given by,

$$\begin{aligned}S_n = \frac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S_5 = \frac{5}{2}[2 \times 3 + (5 - 1) \times 3] \\[1em] = \frac{5}{2}[6 + 4 \times 3] \\[1em] = \frac{5}{2}[6 + 12] \\[1em] = \frac{5}{2} \times 18 \\[1em] = 45.\end{aligned}$$

Hence, Option 1 is the correct option.

Two digit numbers which are divisible by 3 are 12, 15, 18, 21, ..... , 99.

The above series is an A.P. with first term = a = 12 and common difference = d = 15 - 12 = 3.

Let 99 be the nth term, we know that

    a_n = a + (n - 1)d
∴ 99 = 12 + (n - 1) × 3
⇒ 99 = 12 + 3n - 3
⇒ 3n + 9 = 99
⇒ 3n = 90
⇒ n = 30.

Hence, Option 3 is the correct option.

The multiples of 4 that lie between 10 and 250 are 12, 16, 20, ....., 248.

The above series is an A.P. with first term = a = 12 and common difference = d = 16 - 12 = 4.

Let 248 be the nth term, we know that

    a_n = a + (n - 1)d
∴ 248 = 12 + (n - 1) × 4
⇒ 248 = 12 + 4n - 4
⇒ 4n = 248 - 8
⇒ 4n = 240
⇒ n = 60.

Hence, Option 2 is the correct option.

The first 10 even whole numbers are 0, 2, 4, 6, 8, 10, 12, 14, 16 , 18.

Sum of an A.P. is given by,

$$\begin{aligned}S_n = \frac{n}{2}(a + l) \\[1em] \therefore S_{10} = \frac{10}{2}(0 + 18) \\[1em] = 5 \times 18 \\[1em] = 90.\end{aligned}$$

Hence, Option 2 is the correct option.

Given, a = $\frac{1}{8}$ and common ratio = r = $\frac{-\frac{1}{4}}{\frac{1}{8}} = -2.$

We know that

    a_n = ar^{n - 1}

$$\begin{aligned}\therefore a_{11} = \frac{1}{8}(-2)^{11 - 1} \\[1em] = \frac{1}{8}(-2)^{10} \\[1em] = \frac{1}{8} \times 2^{10} \\[1em] = \frac{2^{10}}{2^3} \\[1em] = 2^7 \\[1em] = 128.\end{aligned}$$

Hence, Option 3 is the correct option.

The above series is a G.P. with a = 2 and d = $\frac{6}{2} = 3.$

nth from end = $l\Big(\frac{1}{r}\Big)^{n - 1}$

∴ 5th term from end = $13122 \times \Big(\frac{1}{3}\Big)^{5 - 1}$

$$\begin{aligned}= 13122 \times \frac{1}{3^4} \\[1em] = \frac{13122}{81} \\[1em] = 162.\end{aligned}$$

Hence, Option 1 is the correct option.

In a G.P.,

$$\begin{aligned}\frac{\text{ any term }}{\text{ preceding term }} = \text{ common ratio }. \\[1em] \therefore \frac{2(k + 1)}{k} = \frac{3(k + 1)}{2(k + 1)} \text{ (Eq 1) } \\[1em] \Rightarrow (2k + 2)(2k + 2) = k(3k + 3) \\[1em] \Rightarrow 4k^2 + 4k + 4k + 4 = 3k^2 + 3k \\[1em] \Rightarrow 4k^2 - 3k^2 + 8k - 3k + 4 = 0 \\[1em] \Rightarrow k^2 + 5k + 4 = 0 \\[1em] \Rightarrow k^2 + k + 4k + 4 = 0 \\[1em] \Rightarrow k(k + 1) + 4(k + 1) = 0 \\[1em] \Rightarrow (k + 1)(k + 4) =\end{aligned}$$

But $k + 1$ ≠ 0 as that will not satisfy Eq 1

$$\begin{aligned}\Rightarrow k + 4 = 0 \\[1em] \Rightarrow k = -4.\end{aligned}$$

Hence, Option 2 is the correct option.

Given sequence : -1, -5, -9,......

Difference between 2nd term and 1st term,

⇒ a₂ - a₁ = -5 - (-1)

= -5 + 1 = -4

Difference between 3rd term and 2nd term,

⇒ a₃ - a₂ = -9 - (-5)

= -9 + 5 = -4.

Since, difference between consecutive terms is equal.

Thus, the given sequence is in arithmetic progression with common difference -4.

Thus, Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

Given, T_n = 4 - 2n

⇒ T₁ = 4 - 2 x 1

= 4 - 2 = 2

⇒ T₂ = 4 - 2 x 2

= 4 - 4 = 0

⇒ T₃ = 4 - 2 x 3

= 4 - 6 = -2

⇒ T₄ = 4 - 2 x 4

= 4 - 8 = -4

So, the sequence 2, 0, -2, -4, ..............

After 2nd term, this A.P. will have all terms negative.

So, assertion (A) is true.

Common difference, T_n - T_{n - 1}

= [4 - 2n] - [4 - 2(n - 1)]

= 4 - 2n - [4 - 2n + 2]

= 4 - 2n - 4 + 2n - 2

= -2.

So, reason (R) is true and reason (R) is the correct explanation of assertion (A).

Hence, option 3 is the correct option.

Given sequence: 3, 8, 13, ..............

First term = 3

Difference between second and first term :

⇒ a₂ - a₁ = 8 - 3 = 5

Difference between third and second term :

⇒ a₃ - a₂ = 13 - 8 = 5

Thus, the above sequence is an A.P. with first term (a) = 3 and common difference (d) = 5.

So, reason (R) is true.

By formula,

a_n = a + (n - 1)d

If 53 is a term of this A.P., then

⇒ 53 = 3 + (n - 1)5

⇒ 53 - 3 = (n - 1)5

⇒ 50 = (n - 1)5

⇒ n - 1 = $\frac{50}{5}$

⇒ n - 1 = 10

⇒ n = 10 + 1 = 11.

Thus, 53 is 11th term of the A.P.

So, assertion (A) is true.

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given x, y, z are in A.P.

Common difference, d = difference of two consecutive terms,

⇒ a₂ - a₁ = y - x

⇒ a₃ - a₂ = z - y

If x, y, z are in A.P., then d will be equal, y - x = z - y

⇒ x - z = 2y ..................(1)

If z, y, x are in A.P.

⇒ a₂ - a₁ = y - z

⇒ a₃ - a₂ = x - y

If z, y, x are in A.P., then d will be equal, y - z = x - y

⇒ -z + x = 2y

⇒ z - x = -2y

From equation (1), we can write that z, y, x are in A.P. with negative d.

So, assertion (A) is true.

From above we can conclude that the term in A.P. taken in reverse order also form an A.P.

So, reason (R) is true. And, reason (R) is the correct explanation of assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Let first n natural numbers : 1, 2, 3, ......., n.

The above sequence is an A.P. with first term = 1, common difference = 1 and number of terms = n.

By formula,

Sum of an A.P. = $\frac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$$\begin{aligned}S = \frac{n}{2} \times [2 \times 1 + (n - 1) \times 1] \\[1em] = \frac{n}{2} \times [2 + n - 1] \\[1em] = \frac{n}{2} \times (n + 1).\end{aligned}$$

So, reason (R) is true.

When n = 99.

The sum of first 99 natural numbers

$$\begin{aligned}= \frac{99(99 + 1)}{2}\\[1em] = \frac{99 \times 100}{2}\\[1em] = \frac{9900}{2}\\[1em] = 4950\end{aligned}$$

So, assertion (A) is true and, reason (R) is the correct explanation of assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given, a = -5 and d = -3.

We know that

   a_n = a + (n - 1)d
∴ a₂ = -5 + (2 - 1) × (-3) = -5 + (-3) = -5 - 3 = -8.
   a₃ = -5 + (3 - 1) × (-3) = -5 + 2 × -3 = -5 - 6 = -11.
   a₄ = -5 + (4 - 1) × (-3) = -5 + 3 × -3 = -5 - 9 = -14.

Hence, the first four terms of the A.P. are -5, -8, -11, -14.

(i) Here,

$$\begin{aligned}a_2 - a_1 = \frac{1}{4} - 0 = \frac{1}{4},\\[1em] a_3 - a_2 = \frac{1}{2} - \frac{1}{4} = \frac{2 - 1}{4} = \frac{1}{4}. \\[1em]\end{aligned}$$

i.e. any term - preceding term = $\frac{1}{4}$ = fixed number.

Hence, the given list of numbers forms an A.P.

For the next terms, we have,

$$\begin{aligned}a_5 = a_4 + d = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1, \\[1em] a_6 = a_5 + d = 1 + \frac{1}{4} = \frac{4 + 1}{4} = \frac{5}{4}, \\[1em] a_7 = a_6 + d = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2}.\end{aligned}$$

So, the next three terms are $1, \frac{5}{4}, \frac{3}{2}.$

(ii) Here,

$$\begin{aligned}a_2 - a_1 = \frac{14}{3} - 5 = \frac{14 - 15}{3} = -\frac{1}{3},\\[1em] a_3 - a_2 = \frac{13}{3} - \frac{14}{3} = \frac{13 - 14}{3} = -\frac{1}{3}. \\[1em]\end{aligned}$$

i.e. any term - preceding term = $-\frac{1}{3}$ = fixed number.
Hence, the given list of numbers forms an A.P.

For the next terms, we have,

$$\begin{aligned}a_5 = a_4 + d = 4 + \Big(-\frac{1}{3}\Big) = 4 - \frac{1}{3} = \frac{12 - 1}{3} = \frac{11}{3}, \\[1em] a_6 = a_5 + d = \frac{11}{3} + \Big(-\frac{1}{3}\Big) = \frac{11}{3} - \frac{1}{3} = \frac{10}{3}, \\[1em] a_7 = a_6 + d = \frac{10}{3} + \Big(-\frac{1}{3}\Big) = \frac{10}{3} - \frac{1}{3} = \frac{9}{3} = 3.\end{aligned}$$

So, the next three terms are $\frac{11}{3}, \frac{10}{3}, 3.$

Given, a_n = 6n + 2.

∴ a₁ = 6(1) + 2 = 6 + 2 = 8,
    a₂ = 6(2) + 2 = 12 + 2 = 14.

We know that, d = any term - preceding term

∴ d = a₂ - a₁ = 14 - 8 = 6.

Hence, the common difference of the A.P. is 6.

Here,
a₂ - a₁ = 12 - 9 = 3,
a₃ - a₂ = 15 - 12 = 3.

i.e. any term - preceding term = 3 = fixed number,

Hence, given list of numbers forms an A.P. with first term = a = 9 and common difference = d = 3.

We know that,

a_n = a + (n - 1)d = 9 + (n - 1) × 3 = 9 + 3n - 3 = 6 + 3n.

Since, a_n = 6 + 3n, so

a₁₆ = 6 + 3 × 16 = 6 + 48 = 54.

Hence, the a_n = 3n + 6 and a₁₆ = 54.

Given, a = 17, l = -40 and d = 14 - 17 = -3.

nth term from the end = l - (n - 1)d

∴ 6th term from the end = -40 - (6 - 1) × (-3) = -40 - 5 × -3 = -40 + 15 = -25.

Hence, 6th term from the end is -25.

Given, a₈ = 31 and a₁₅ - a₁₁ = 16.

We know that

⇒ a_n = a + (n - 1)d
∴ a₈ = a + 7d      (Eq 1)
   a₁₅ = a + 14d and
   a₁₁ = a + 10d.

Given, a₁₅ - a₁₁ = 16.
∴ a + 14d - (a + 10d) = 16
⇒ a - a + 14d - 10d = 16
⇒ 4d = 16
⇒ d = 4.

Putting value of d in Eq 1
⇒ a₈ = a + 7d
⇒ a + 7 × 4 = 31
⇒ a + 28 = 31
⇒ a = 31 - 28
⇒ a = 3.

Hence, the terms of A.P. are

a₂ = a₁ + d = 3 + 4 = 7,
a₃ = a₂ + d = 7 + 4 = 11,
a₄ = a₃ + d = 11 + 4 = 15.

Hence, the terms of the A.P. are 3, 7, 11, 15, ....

We know that

a_n = a + (n - 1)d

According to question,

⇒ a₁₇ = 2(a₈) + 5
⇒ a + 16d = 2(a + 7d) + 5
⇒ a + 16d = 2a + 14d + 5
⇒ 2a - a + 14d - 16d + 5 = 0
⇒ a - 2d + 5 = 0
⇒ a = 2d - 5.     (Eq 1)

Given, a₁₁ = 43

⇒ a + 10d = 43
⇒ 2d - 5 + 10d = 43
⇒ 12d = 43 + 5
⇒ 12d = 48
⇒ d = 4.

Putting value of d in Eq 1,

⇒ a = 2d - 5
⇒ a = 2 × 4 - 5 = 8 - 5 = 3.

∴ a_n = 3 + (n - 1) × 4 = 3 + 4n - 4 = 4n - 1.

Hence, the nth term of the A.P. is 4n - 1.

We know that

    a_n = a + (n - 1)d

According to question,

⇒ a₁₉ = 3(a₆)
⇒ a + 18d = 3(a + 5d)
⇒ a + 18d = 3a + 15d
⇒ 3a - a = 18d - 15d
⇒ 2a = 3d
⇒ a = $\frac{3}{2}d$     (Eq 1)

Given, a₉ = 19

$$\begin{aligned}\Rightarrow a + 8d = 19 \\[1em] \Rightarrow \frac{3}{2}d + 8d = 19 \\[1em] \Rightarrow \frac{3d + 16d}{2} = 19 \\[1em] \Rightarrow \frac{19d}{2} = 19 \\[1em] \Rightarrow d = \frac{19}{19} \times 2 \\[1em] \Rightarrow d = 2. \\[1em]\end{aligned}$$

Putting value of d in Eq 1 we get,

$$\begin{aligned}\Rightarrow a = \frac{3}{2}d \\[1em] \Rightarrow a = \frac{3}{2} \times 2 \\[1em] \Rightarrow a = 3.\end{aligned}$$

Hence, the terms of A.P. are

a₂ = a₁ + d = 3 + 2 = 5,
a₃ = a₂ + d = 5 + 2 = 7,
a₄ = a₃ + d = 7 + 2 = 9.

Hence, the terms of the A.P. are 3, 5, 7, 9, ....

We know that

    a_n = a + (n - 1)d

Given,

    a₃ = 4 and a₉ = -8
∴ a + 2d = 4 and a + 8d = -8

Subtracting the two Equations,

⇒ a + 8d - (a + 2d) = -8 - 4
⇒ a - a + 8d - 2d = -12
⇒ 6d = -12
⇒ d = -2.

Putting value of d in a + 2d = 4

⇒ a + 2 × (-2) = 4
⇒ a - 4 = 4
⇒ a = 4 + 4
⇒ a = 8.

Let nth term of the A.P. be zero so, a_n = 0.

⇒ a + (n - 1)d = 0
⇒ 8 + (n - 1) × (-2) = 0
⇒ 8 - 2n + 2 = 0
⇒ 10 - 2n = 0
⇒ 2n = 10
⇒ n = 5.

Hence, the 5th term of the A.P. is zero.

The above list is an A.P. with a = 5 and d = 2 - 5 = -3.

Let the nth term of the A.P. be -55 or, a_n = -55.

We know that

a_n = a + (n - 1)d

⇒ -55 = 5 + (n - 1) × (-3)
⇒ -55 = 5 - 3n + 3
⇒ -55 = 8 - 3n
⇒ 3n = 8 + 55
⇒ 3n = 63
⇒ n = 21.

Hence, the 21st term of the A.P. is -55..

We know that

a_n = a + (n - 1)d

Given,
a₂₄ = 2(a₁₀)
∴ a + 23d = 2(a + 9d)
⇒ a + 23d = 2a + 18d
⇒ 2a - a = 23d - 18d
⇒ a = 5d.

By using formula for nth term and a = 5d

a₇₂ = a + 71d = 5d + 71d = 76d
a₁₅ = a + 14d = 5d + 14d = 19d

∴ 4a₁₅ = 4 × 19d = 76d = a₇₂.

Hence, proved that 72nd term is four times the 15th term of the A.P.

The above series is an A.P. with a = 20 and

$$\begin{aligned}d = 19\frac{1}{4} - 20 \\[1em] = \frac{77}{4} - 20 \\[1em] = \frac{77 - 80}{4} \\[1em] = -\frac{3}{4}.\end{aligned}$$

Let nth term be the first negative number.

We know that

a_n = a + (n - 1)d

$$\begin{aligned}\therefore a_n = 20 + (n - 1) \times -\frac{3}{4} \\[1em] = 20 - \frac{3}{4}n + \frac{3}{4} \\[1em] = \frac{80 + 3}{4} - \frac{3}{4}n \\[1em] = \frac{83}{4} - \frac{3}{4}n \\[1em]\end{aligned}$$

Since a_n is a negative number

$$\begin{aligned}\therefore a_n \lt 0 \\[1em] \Rightarrow \frac{83}{4} - \frac{3}{4}n \lt 0 \\[1em] \Rightarrow \frac{3}{4}n \gt \frac{83}{4} \\[1em] \Rightarrow n \gt \frac{83}{4} \times \frac{4}{3} \\[1em] \Rightarrow n \gt \frac{83}{3} \\[1em] \Rightarrow n \gt 27\frac{2}{3} \\[1em] \therefore n = 28.\end{aligned}$$

Hence, 28th term is the first negative number in the A.P.

The three digit numbers that are divisible by 9 are 108, 117, 126, ....., 999.

The above series is in A.P. with a = 108, d = 117 - 108 = 9 and l = 999.

Let the nth term be last term

∴ a_n = 999.

We know that

a_n = a + (n - 1)d

⇒ 999 = 108 + (n - 1) × 9
⇒ 108 + 9n - 9 = 999
⇒ 9n + 99 = 999
⇒ 9n = 900
⇒ n = 100.

Hence, 100 three digit numbers are divisible by 9.

Let the three numbers in A.P. be a - d, a, a + d.

Given, S_n = -3

⇒ a - d + a + a + d = -3
⇒ 3a = -3
⇒ a = -1.

Given, product of numbers = 8

⇒ (a - d)a(a + d) = 8

Putting value of a = -1

⇒ (-1 - d)(-1)(-1 + d) = 8
⇒ (d + 1)(d - 1) = 8
⇒ d² - 1 = 8
⇒ d² = 9
⇒ d = -3, 3.

Putting d = -3
Numbers ⇒ a - d = -1 - (-3) = -1 + 3 = 2, a = -1, a + d = -1 + (-3) = -4.

Putting d = 3
Numbers ⇒ a - d = -1 - 3 = -4, a = -1, a + d = -1 + 3 = 2.

Hence, the three numbers in A.P. are -4, -1, 2 or 2, -1, -4.

Let the angles of quadrilateral be a, (a + d), (a + 2d), (a + 3d).

According to question greatest angle is twice the smallest angle,

⇒ (a + 3d) = 2a
⇒ 2a - a = 3d
⇒ a = 3d.

By putting values of a = 3d , the angles become,

a = 3d

a + d = 3d + d = 4d

a + 2d = 3d + 2d = 5d

a + 3d = 3d + 3d = 6d

Sum of angles of quadrilateral = 360°.

∴ 3d + 4d + 5d + 6d = 360°
⇒ 18d = 360°
⇒ d = 20°.

Hence, the angles are

⇒ 3d = 3 × 20° = 60°

⇒ 4d = 4 × 20° = 80°

⇒ 5d = 5 × 20° = 100°

⇒ 6d = 6 × 20° = 120°.

Hence, the angles of the quadrilateral are 60°, 80°, 100°, 120°.

Since, a_n = 15 - 4n

∴ a₁ = 15 - 4(1) = 15 - 4 = 11 and a₂₀ = 15 - 4(20) = 15 - 80 = -65.

We know that

$$\begin{aligned}S_n = \frac{n}{2}[a + l] \\[1em] \therefore S_{20} = \frac{20}{2}[11 + (-65)] \\[1em] = 10[-54] \\[1em] = -540.\end{aligned}$$

Hence, the sum of first 20 terms of an A.P. is -540.

Total number of students (n) = 10

Height of first student (a) = 150 cm

Difference between height of every next student (d) = 2 cm

Since, the height difference between every next student is equal, so it can be considered an A.P.

By formula,

Sum of A.P. (S) = $\frac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$$\begin{aligned}S = \frac{10}{2} \times [2 \times 150 + (10 - 1) \times 2] \\[1em] = 5 \times [300 + 9 \times 2] \\[1em] = 5 \times [300 + 18] \\[1em] = 5 \times 318 \\[1em] = 1590 \text{ cm}.\end{aligned}$$

Hence, the total sum of heights of ten students = 1590 cm.

Given, a₄ = 54 and a₇ = 1458.

We know that in G.P.

   a_n = ar^{n - 1}
∴ a₄ = ar³ and a₇ = ar⁶.

Dividing a₇ by a₄,

$$\begin{aligned}\Rightarrow \frac{ar^6}{ar^3} = \frac{1458}{54} \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r = \sqrt[3]{27} \\[1em] \therefore r = 3. \\[1em]\end{aligned}$$

Putting value of r in ar³ = 54,

$$\begin{aligned}\Rightarrow a(3)^3 = 54 \\[1em] \Rightarrow 27a = 54 \\[1em] \Rightarrow a = 2. \\[1em]\end{aligned}$$

Terms of G.P. are

⇒ a₂ = ar = 2 × 3 = 6,
⇒ a₃ = ar² = 2 × 3² = 18,
⇒ a₄ = ar³ = 2 × 3³ = 54.

Hence, the G.P. is 2, 6, 18, 54, ....

Given, a₄ = (a₂)² and a = -3.

   a_n = ar^{n - 1}
∴ a₄ = ar³ and a₂ = ar.

Since, a₄ = (a₂)²
∴ ar³ = (ar)²
⇒ (-3)r³ = (-3r)²
⇒ -3r³ = 9r²
⇒ 9r² + 3r³ = 0
⇒ r²(9 + 3r) = 0
⇒ r² = 0 or 9 + 3r = 0
⇒ r = 0 or r = -3.

As common ratio cannot be equal to zero so, r ≠ 0.

a₇ = ar⁶ = (-3)(-3)⁶ = -3 × 729 = -2187.

Hence, the 7th term of the G.P. is -2187.

Given, a₄ = x, a₁₀ = y and a₁₆ = z.

We know that
   a_n = ar^{n - 1}
∴ a₄ = ar³ = x, a₁₀ = ar⁹ = y and a₁₆ = ar¹⁵ = z.

Dividing y by x we get,

$\frac{y}{x} = \frac{ar^9}{ar^3} = r^6$

Dividing z by y we get,

$$\begin{aligned}\frac{z}{y} = \frac{ar^{15}}{ar^9} = r^6 \\[1em] \therefore \frac{y}{z} = \frac{z}{y}.\end{aligned}$$

Hence, proved that x, y, z are in G.P.

Here, a = 3 and r = $\frac{\frac{3}{2}}{3} = \frac{3}{2 \times 3} = \frac{1}{2}.$

Let n terms are needed for the sum.

$$\begin{aligned}S_n = \frac{a(r^n - 1)}{r - 1} \\[1em] \Rightarrow \frac{3069}{512} = \frac{3\Big[\Big(\frac{1}{2}\Big)^n - 1\Big]}{\frac{1}{2} - 1} \\[1em] \Rightarrow \frac{3069}{512} = \frac{3\Big[\Big(\frac{1}{2}\Big)^n - 1\Big]}{\frac{1 - 2}{2}} \\[1em] \Rightarrow \frac{3069}{512} = -6\Big[\Big(\frac{1}{2}\Big)^n - 1\Big] \\[1em] \Rightarrow \frac{3069}{512 \times 6} = 1 - \Big(\frac{1}{2}\Big)^n \\[1em] \Rightarrow \Big(\frac{1}{2}\Big)^n = 1 - \frac{3069}{3072} \\[1em] \Rightarrow \Big(\frac{1}{2}\Big)^n = \frac{3072 - 3069}{3072} \\[1em] \Rightarrow \Big(\frac{1}{2}\Big)^n = \frac{3}{3072} \\[1em] \Rightarrow \Big(\frac{1}{2}\Big)^n = \frac{1}{1024} \\[1em] \Rightarrow \Big(\frac{1}{2}\Big)^n = \Big(\frac{1}{2}\Big)^{10} \\[1em] \therefore n = 10.\end{aligned}$$

Hence, 10 terms of the G.P. are required for the sum of $\frac{3069}{512}$.

Given,

G.P. : 15, 30, 60, 120 ......

First term (a) = 15

Common ratio (r) = $\frac{30}{15}$ = 2

(a) nth term of G.P. = ar^{n - 1} = 15 x 2^{n - 1}

= $\frac{15}{2}$ x 2ⁿ

= 7.5 x 2ⁿ

Hence, nth term of the given G.P. is 7.5 x 2ⁿ

(b) Let sum of n terms of G.P. is 945.

By formula,

Sum of n terms of G.P. = $\frac{a(r^n - 1)}{(r - 1)}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 945 = \frac{15.(2^n - 1)}{2 - 1} \\[1em] \Rightarrow 945 = \frac{15.(2^n - 1)}{1} \\[1em] \Rightarrow 2^n - 1 = \frac{945}{15} \\[1em] \Rightarrow 2^n - 1 = 63 \\[1em] \Rightarrow 2^n = 63 + 1 \\[1em] \Rightarrow 2^n = 64 \\[1em] \Rightarrow 2^n = 2^6 \\[1em] \Rightarrow n = 6.\end{aligned}$$

Hence, sum of 6 terms of G.P. = 945.

Given,

The roots of the equation (q - r)x² + (r - p)x + (p - q) = 0 are equal.

∴ Discriminant (D) = 0

⇒ b² - 4ac = 0

⇒ (r - p)² - 4 × (q - r) × (p - q) = 0

⇒ r² + p² - 2pr - 4(qp - q² - rp + qr) = 0

⇒ r² + p² - 2pr - 4qp + 4q² + 4rp - 4qr = 0

⇒ r² + p² + 2pr - 4qp - 4qr + 4q² = 0

⇒ (p + r)² - 4q(p + r) + 4q² = 0

Let p + r = y

⇒ y² - 4qy + 4q² = 0

⇒ (y - 2q)² = 0

⇒ y - 2q = 0

⇒ y = 2q

⇒ p + r = 2q.

Hence, proved that p + r = 2q.