Arithmetic Progression

Since, 13 - 11 = 2, 15 - 13 = 2, 17 - 15 = 2 and 19 - 17 = 2.

Hence, the series is an A.P. with common difference = 2.

We know that n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

⇒ a₂₀ = 11 + (20 - 1)2

= 11 + (19)2

= 11 + 38

= 49.

Hence, a = 11, d = 2 and a₂₀ = 49.

Since, 20 - 13 = 7, 27 - 20 = 7, and 34 - 27 = 7.

Hence, the series is an A.P. with common difference = 7.

We know that,

n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

⇒ a₁₆ = 13 + (16 - 1)7

= 13 + (15) × 7

= 13 + 105

= 118.

Hence, a = 13, d = 7 and a₁₆ = 118.

Since, 6 - 10 = -4, 2 - 6 = -4, -2 - 2 = -4 and -6 - (-2) = -4.

Hence, the series is an A.P. with common difference = -4.

We know that n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

⇒ a₁₀ = 10 + (10 - 1).(-4)

= 10 + (9)-4

= 10 - 36

= -26.

Hence, a = 10, d = -4 and a₂₀ = -26.

Since, 5 - 11 = -6, -1 - 5 = -6, -7 - (-1) = -6 and -13 - (-7) = -6.

Hence, the series is an A.P. with common difference = -6.

We know that n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

⇒ a₁₂ = 11 + (12 - 1).(-6)

= 11 + 11(-6)

= 11 - 66

= -55.

Hence, a = 11, d = -6 and a₁₂ = -55.

$$\begin{aligned}7\frac{3}{4} = \frac{31}{4} = 7.75 \\[1em] 9\frac{1}{2} = \frac{19}{2} = 9.5 \\[1em] 11\frac{1}{4} = \frac{45}{4} = 11.25 \\[1em]\end{aligned}$$

Since, 7.75 - 6 = 1.75, 9.5 - 7.75= 1.75 and 11.25 - 9.5 = 1.75.

Hence, the series is an A.P. with common difference = 1.75.

We know that n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

⇒ a₁₇ = 6 + (17 - 1)1.75

= 6 + (16)1.75

= 6 + 28

= 34.

Hence, a = 6, d = $\frac{7}{4}$ and a₁₇ = 34.

Series :

$\Rightarrow 6, 5\frac{1}{2}, 5, 4\frac{1}{2}, 4$

⇒ 6, 5.5, 5, 4.5, 4

Since, 5.5 - 6 = -0.5, 5 - 5.5 = -0.5, 4.5 - 5 = -0.5 and 4.0 - 4.5 = -0.5.

Hence, the series is an A.P. with common difference (d) = -0.5

We know that n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

⇒ a₉ = 6 + (9 - 1)(-0.5)

= 6 + 8(-0.5)

= 6 - 4

= 2.

Hence, a = 6, d = -0.5 and a₉ = 2.

Given,

a = 5

d = 11 - 5 = 6

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

a_n = 5 + (n - 1)6

= 5 + 6n - 6

= 6n - 1.

(i) 11th term:

⇒ a₁₁ = 5 + (11 - 1)6

= 5 + 10(6)

= 5 + 60

= 65.

(ii) 21st term

⇒ a₂₁ = 5 + (21 - 1)6

= 5 + (20)6

= 5 + 120

= 125.

Hence, nth term = 6n - 1, 11th term = 65 and 21st term = 125.

Given,

a = 13

d = 7 - 13 = -6

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

a_n = 13 + (n - 1)(-6)

= 13 - 6n + 6

= 19 - 6n.

(i) 9th term

⇒ a₉ = 13 + (9 - 1).(-6)

= 13 + 8(-6)

= 13 - 48

= -35.

(ii) 16th term

⇒ a₁₆ = 13 + (16 - 1).(-6)

= 13 + (15).(-6)

= 13 - 90

= -77.

Hence, nth term = 19 - 6n, 9th term = -35 and 16th term = -77.

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given,

⇒ a = 6

⇒ d = 10 - 6 = 4

⇒ a_n = 174

⇒ 174 = 6 + (n - 1)4

⇒ 174 - 6 = (n - 1)4

⇒ 168 = (n - 1)4

⇒ $\frac{168}{4}$ = n - 1

⇒ 42 = n - 1

⇒ n = 42 + 1

⇒ n = 43.

Hence, there are 43 terms in the A.P.

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given,

a = 41

d = 38 - 41 = -3

Let nth term be -1.

⇒ a_n = -1

⇒ a + (n - 1)d = -1

⇒ -1 = 41 + (n - 1)(-3)

⇒ - 1 - 41 = (n - 1)(-3)

⇒ -42 = (n - 1)(-3)

⇒ $\frac{-42}{-3}$ = n - 1

⇒ 14 = n - 1

⇒ n = 14 + 1

⇒ n = 15.

Hence, there are 15 terms in the A.P.

In the A.P. 3, 8, 13, 18, 23,...., a = 3 and d = 8 - 3 = 5.

Let n^th term be 98.

∴ a_n = 98

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

⇒ 98 = 3 + (n - 1)5

⇒ 98 - 3 = (n - 1)5

⇒ 95 = (n - 1)5

⇒ (n - 1) = $\frac{95}{5}$

⇒ n - 1 = 19

⇒ n = 19 + 1

⇒ n = 20.

Hence, 20th term of A.P. is 98.

In the A.P. 72, 68, 64, 60,...., a = 72 and d = 68 - 72 = -4.

Let n^th term be 0.

∴ a_n = 0

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

⇒ 0 = 72 + (n - 1)(-4)

⇒ -72 = (n - 1)(-4)

⇒ n - 1 = $\frac{-72}{-4}$

⇒ n - 1 = 18

⇒ n = 18 + 1

⇒ n = 19.

Hence, 19th term of A.P. is 0.

Series :

$\Rightarrow 1, 1\frac{1}{6}, 1\frac{1}{3}, ......$

$\Rightarrow 1, \frac{7}{6}, \frac{4}{3}, ......$

In the A.P. 1, $1\frac{1}{6}, 1\frac{1}{3}$,...., a = 1 and d = $\frac{7}{6} - 1 = \frac{1}{6}$

Let n^th term be 3.

∴ a_n = 3

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

$$\begin{aligned}\Rightarrow 3 = 1 + (n - 1)\frac{1}{6} \\[1em] \Rightarrow 3 = 1 + (n - 1)\frac{1}{6} \\[1em] \Rightarrow 3 - 1 = (n - 1)\frac{1}{6} \\[1em] \Rightarrow 2 = (n - 1)\frac{1}{6} \\[1em] \Rightarrow 2 \times 6 = (n - 1) \\[1em] \Rightarrow 12 = (n - 1) \\[1em] \Rightarrow n = 12 + 1 \\[1em] \Rightarrow n = 13.\end{aligned}$$

Hence, 13th term of A.P. is 3.

We know that,

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given, 4th term is 13.

∴ a₄ = a + (4 - 1)d

⇒ 13 = a + 3d

⇒ a + 3d = 13 .......(1)

Given, 10th term is 25

∴ a₁₀ = a + (10 - 1)d

⇒ 25 = a + 9d

⇒ a + 9d = 25 ......(2)

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

⇒ a + 9d - (a + 3d) = 25 - 13

⇒ a + 9d - a - 3d = 12

⇒ 6d = 12

⇒ d = $\frac{12}{6}$

⇒ d = 2.

Substituting value of d in (1) we get,

⇒ a + 3(2) = 13

⇒ a + 6 = 13

⇒ a = 13 - 6

⇒ a = 7.

∴ a₁₇ = 7 + (17 - 1)2

= 7 + (16)2

= 7 + 32

= 39.

Hence, a = 7, d = 2 and a₁₇ = 39.

Let first term be a and common difference be d.

We know that,

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given, 7th term is -4.

∴ a₇ = a + (7 - 1)d

⇒ -4 = a + 6d

⇒ a + 6d = -4 ....(1)

Given,

13th term is -16

∴ a₁₃ = a + (13 - 1)d

⇒ -16 = a + 12d

⇒ a + 12d = -16 ....(2)

Subtracting (1) from (2) we get,

⇒ a + 12d - (a + 6d) = -16 - (-4)

⇒ a + 12d - a - 6d = -16 + 4

⇒ 6d = -12

⇒ d = $\frac{-12}{6}$

⇒ d = -2.

Substituting value of d in (1) we get,

⇒ a + 6(-2) = -4

⇒ a - 12 = -4

⇒ a = -4 + 12

⇒ a = 8.

∴ a_n = 8 + (n - 1)(-2)

= 8 - 2n + 2

= 10 - 2n.

Hence, a = 8, d = -2 and a_n = 10 - 2n.

We know that,

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given,

a = 7

d = 10 - 7 = 3

⇒ a_n = 184

⇒ a_n = 7 + (n - 1)3

⇒ 184 = 7 + (n - 1)3

⇒ 184 - 7 = (n -1)3

⇒ 177 = (n - 1)3

⇒ $\frac{177}{3}$ = (n - 1)

⇒ (n - 1) = 59

⇒ n = 59 + 1

⇒ n = 60.

There are 60 terms in the A.P.

The 8th term from end of the A.P. is 53rd (60 - 7) term from the starting.

⇒ a₅₃ = 7 + (53 - 1)3

= 7 + (52)3

= 7 + 156

= 163.

Hence, the 8th term from end of the A.P is 163.

We know that,

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given,

a = 17

d = 14 - 17 = -3

Let no. of terms be n.

⇒ a_n = -40

By formula,

⇒ a_n = a + (n - 1)d

⇒ -40 = 17 + (n - 1)(-3)

⇒ -40 - 17 = (n - 1)(-3)

⇒ -57 = (n - 1)(-3)

⇒ $\frac{-57}{-3}$ = n - 1

⇒ 19 = n - 1

⇒ n = 19 + 1

⇒ n = 20.

There are 20 terms in the A.P.

The 8th term from end of the A.P. = 20 - 5 = 15

⇒ a₁₅ = 17 + (15 - 1)-3

= 17 + (14)-3

= 17 - 42

= -25

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

Since, (5x + 2), (4x - 1) and (x + 2) are in A.P.

Hence, difference between consecutive terms are equal.

∴ (4x - 1) - (5x + 2) = (x + 2) - (4x - 1)

⇒ 4x - 1 - 5x - 2 = x + 2 - 4x + 1

⇒ -3 - x = 3 - 3x

⇒ -3 - 3 = -3x + x

⇒ -6 = -2x

⇒ x = $\frac{-6}{-2}$

⇒ x = 3.

Hence, the value of x = 3.

Since, (k - 3), (2k + 1) and (4k + 3) are in A.P.

Hence, difference between consecutive terms are equal.

∴ 2k + 1 - (k - 3) = 4k + 3 - (2k + 1)

⇒ 2k + 1 - k + 3 = 4k + 3 - 2k - 1

⇒ k + 4 = 2k + 2

⇒ 4 - 2 = 2k - k

⇒ k = 2.

Hence, the value of k = 2.

Let the three numbers in the Arithmetic Progression (A.P.) be a - d, a, a + d.

Given,

The sum of the three numbers is 15.

⇒ a - d + a + a + d = 15

⇒ 3a = 15

⇒ a = $\frac{15}{3}$

⇒ a = 5.

The middle term of the A.P. is 5.

Given,

The product of the three numbers is 80.

⇒ (a - d)(a)(a + d) = 80

⇒ (5 - d)(5)(5 + d) = 80

⇒ (5 - d)(5 + d) = $\frac{80}{5}$

⇒ (5 - d)(5 + d) = 16

⇒ 5² - d² = 16

⇒ 25 - d² = 16

⇒ d² = 25 - 16

⇒ d² = 9

⇒ d = $\sqrt{9}$

⇒ d = 3 or d = -3.

Case 1 : If a = 5 and d = 3

⇒ a - d = 5 - 3 = 2

⇒ a = 5

⇒ a + d = 5 + 3 = 8.

Case 2 : If a = 5 and d = -3

⇒ a - d = 5 - (-3) = 8

⇒ a = 5

⇒ a + d = 5 + (-3) = 2.

Hence, the numbers are 2, 5, 8.

Let the four angles of the quadrilateral be in A.P. with common difference 10° be:

a, a + 10°, a + 20°, a + 30°

The sum of the interior angles of any quadrilateral is always 360°.

⇒ a + a + 10° + a + 20° + a + 30° = 360°

⇒ 4a + 60° = 360°

⇒ 4a = 360° - 60°

⇒ 4a = 300°

⇒ a = $\frac{300°}{4}$

⇒ a = 75°

⇒ a + 10° = 75° + 10° = 85°

⇒ a + 20° = 75° + 20° = 95°

⇒ a + 30° = 75° + 30° = 105°.

Hence, the angles of quadrilateral are 75°, 85°, 95°, 105°.

Let the three angles of triangle in A.P. be represented as : a - d, a, a + d.

The sum of the interior angles of any triangle is always 180°.

⇒ a - d + a + a + d = 180°

⇒ 3a = 180

⇒ a = $\frac{180}{3}$

⇒ a = 60°.

Given,

The common difference is 20°

⇒ a - 20° = 60° - 20° = 40°

⇒ a + 20° = 60° + 20° = 80°.

Hence, the angles of triangle are 40°, 60°, 80°.

(i) Given,

-11, -7, -3, ......,49, 53

Second term - First term : −7 − (−11) = 4

Third term - Second term : −3 − (−7) = 4

Last term - Second last term : 53 − 49 = 4

Since the difference between consecutive terms is constant, the progression is an Arithmetic Progression (AP) with first term (a) = -11 and common difference (d) = 4.

Hence, the progression is an Arithmetic Progression.

(ii) Let 53 be the nth term.

By formula,

⇒ t_n = a + (n - 1)d

⇒ 53 = -11 + (n - 1)4

⇒ 53 = -11 + 4n - 4

⇒ 53 = -15 + 4n

⇒ 4n = 53 + 15

⇒ 4n = 68

⇒ n = $\frac{68}{4}$

⇒ n = 17.

Hence, there are a total of 17 terms in the progression.

(iii) Since, n = 17, is odd.

By formula,

Middle term = $\frac{n + 1}{2}$

= $\frac{17 + 1}{2}$

= $\frac{18}{2}$

= 9.

Thus, 9th term is middle term.

By formula,

⇒ t₉ = a + (n - 1)d

= -11 + (9 - 1) × 4

= -11 + 8 × 4

= -11 + 32

= 21.

Hence, the value of the middle most term is 21.

The above sequence is an A.P. with common difference = 12 - 7 = 5.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

⇒ S₁₉ = $\frac{19}{2}$ [2(2) + (19 - 1)5]

= 9.5 [4 + (18)5]

= 9.5 [4 + 90]

= 9.5 (94)

= 893.

Hence, S₁₉ = 893.

The above sequence is an A.P. with common difference = 5 - 7 = -2.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

⇒ S₁₄ = $\frac{14}{2}$ [2(9) + (14 - 1)(-2)]

= 7[18 + (13)(-2)]

= 7[18 - 26]

= 7.(-8)

= -56.

Hence, S₁₄ = -56.

The above sequence is an A.P. with common difference = -7 - (-11) = 4.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

⇒ S₁₂ = $\frac{12}{2}$ [2(-11) + (12 - 1)4]

= 6.[-22 + (11)4]

= 6.[-22 + 44]

= 6.(22)

= 132.

Hence, S₁₂ = 132.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

Given,

a = $\frac{1}{15}$

d = $\frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}$

n = 11

$$\begin{aligned}\Rightarrow S_{11} = \frac{11}{2} \Big[2\Big(\frac{1}{15}\Big) + (11 - 1)\Big(\frac{1}{60}\Big)\Big] \\[1em] = \frac{11}{2} \Big[\Big(\frac{2}{15}\Big) + \Big(\frac{10}{60}\Big)\Big] \\[1em] = \frac{11}{2} \Big[\frac{2 \times 2 + 1 \times 5 }{30}\Big] \\[1em] = \frac{11}{2} \Big[\frac{4 + 5}{30}\Big] \\[1em] = \frac{11}{2} \Big[\frac{9}{30}\Big] \\[1em] = \frac{11}{2} \Big[\frac{3}{10}\Big] \\[1em] = \frac{33}{20}.\end{aligned}$$

Hence, S₁₁ = $\frac{33}{20}$.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

Given,

a = 0.6

d = 1.7 - 0.6 = 1.1

n = 100

⇒ S₁₀₀ = $\frac{100}{2}$ [2(0.6) + (100 - 1)1.1]

= 50 [1.2 + (99)1.1]

= 50 [1.2 + 108.9]

= 50 (110.1)

= 5505.

Hence, S₁₀₀ = 5505.

Series :

7 + 10.5 + 14 + ....... + 84

The above series is an A.P. with common difference = 3.5

Let 84 be nth term in the A.P.

We know that,

∴ a_n = a + (n - 1)d

⇒ 84 = 7 + (n - 1) × 3.5

⇒ 84 - 7 = (n - 1) × 3.5

⇒ 77 = (n - 1) × 3.5

⇒ $\frac{77}{3.5}$ = (n - 1)

⇒ 22 = (n - 1)

⇒ n = 22 + 1

⇒ n = 23.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ (a + l)

⇒ S₂₃ = $\frac{23}{2}$ (7 + 84)

= $\frac{23}{2}$ (91)

= $\frac{2093}{2} = 1046\frac{1}{2}$

Hence, S₂₃ = $1046\frac{1}{2}$.

Let total number of terms in the A.P. be n.

We know that,

∴ a_n = a + (n - 1)d

⇒ 10 = 32 + (n - 1).(-2)

⇒ 10 - 32 = (n - 1).(-2)

⇒ -22 = (n - 1).(-2)

⇒ $\frac{-22}{-2}$ = (n - 1)

⇒ 11 = n - 1

⇒ n = 11 + 1

⇒ n = 12.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ (a + l)

⇒ S₁₂ = $\frac{12}{2}$ (32 + 10)

= 6 × 42

= 252.

Hence, S₁₂ = 252.

Let total number of terms in an A.P. be n.

We know that,

∴ a_n = a + (n - 1)d

Given,

a = -5

d = -8 - (-5) = -3

⇒ a_n = -5 + (n - 1)-3

⇒ -62 = -5 + (n - 1)-3

⇒ -62 - (-5) = (n - 1)-3

⇒ -62 - (-5) = (n - 1)-3

⇒ -57 = (n - 1)-3

⇒ $\frac{-57}{-3}$ = (n - 1)

⇒ 19 = (n - 1)

⇒ n = 19 + 1

⇒ n = 20

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ (a + l)

⇒ S₂₀ = $\frac{20}{2}$ [-5 + (-62)]

= 10 [-5 + (-62)]

= 10(-67)

= -670.

Hence, S₂₀ = -670.

The 2-digit natural numbers divisible by 5 form an A.P., with common difference = 5.

10, 15, 20,.....,95

Let total number of terms in an A.P. be n.

We know that,

∴ a_n = a + (n - 1)d

⇒ 95 = 10 + (n - 1)5

⇒ 95 - 10 = (n - 1)5

⇒ 85 = (n - 1)5

⇒ $\frac{85}{5}$ = (n - 1)

⇒ 17 = (n - 1)

⇒ n = 17 + 1

⇒ n = 18.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ (a + l)

⇒ S₁₈ = $\frac{18}{2} \times (10 + 95)$

= 9(105)

= 945.

Hence, S₁₈ = 945.

Even numbers between 10 and 100 are :

12, 14, 16, ....., 98

Let total number of terms in A.P. be n.

Given,

a = 12

d = 14 - 12 = 2

We know that,

∴ a_n = a + (n - 1)d

⇒ 98 = 12 + (n - 1)2

⇒ 98 - 12 = (n - 1)2

⇒ 86 = (n - 1)2

⇒ $\frac{86}{2}$ = (n - 1)

⇒ n - 1 = 43

⇒ n = 43 + 1

⇒ n = 44

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ (a + l)

⇒ S₄₄ = $\frac{44}{2}$ (12 + 98)

= 22 (110)

= 2420.

Hence, S₄₄ = 2420.

The odd numbers between 100 and 150 are :

101, 103, 105,…….., 149.

Let total number of terms in A.P. be n.

We know that,

∴ an = a + (n - 1)d

Given,

a = 101

an = 149

d = 103 - 101 = 2

⇒ 149 = 101 + (n - 1)2

⇒ 149 - 101 = (n - 1)2

⇒ 48 = (n - 1)2

⇒ $\frac{48}{2}$ = (n - 1)

⇒ 24 = (n - 1)

⇒ n = 24 + 1

⇒ n = 25.

We know that,

Sum of n terms of an A.P. is given by,

S_n = $\frac{n}{2}$ (a + l)

⇒ S₂₅ = $\frac{25}{2}$ (101 + 149)

⇒ $\frac{25}{2}$ ×(250)

= 25 × 125

= 3125.

Hence, S₂₅ = 3125.

8, 16, 24,......,(upto 15th term).

The above is an A.P. with common difference equal to 8.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

Given,

a = 8

d = 16 - 8 = 8

n = 15

S₁₅ = $\frac{15}{2}$ [2(8) + (15 - 1)8]

= $\frac{15}{2} \times$ [16 + (14)8]

= $\frac{15}{2} \times$ [16 + 112]

= $\frac{15}{2} \times$ (128)

= 15 × 64

= 960.

Hence, S₁₅ = 960.

T_n = 6(7 - n)

Substituting n = 1, we get :

T₁ = 6(7 - 1) = 6 × 6 = 36.

Substituting n = 2, we get :

T₂ = 6(7 - 2) = 6 × 5 = 30.

Common difference (d) = T₂ - T₁

= 30 - 36 = -6.

Hence, common difference (d) = -6 and first term = 36.

(b) By formula,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

= $\frac{25}{2}$ [2 × 36 + (25 − 1) × −6]

= 12.5 × [72 + 24× − 6]

= 12.5 × [72 − 144]

= 12.5 × −72

= −900.

Hence, sum of its first 25 terms = -900.

Let a be the first term and d be the common difference.

We know that,

∴ a_n = a + (n - 1)d

Given,

The 4th term of an A.P. is 22.

⇒ a₄ = a + (4 - 1)d

⇒ 22 = a + 3d

⇒ a + 3d = 22 .....(1)

Given,

The 15th term of an A.P. is 66.

⇒ a₁₅ = a + (15 - 1)d

⇒ 66 = a + 14d

⇒ a + 14d = 66 .....(2)

Subtracting Equation 1 from Equation 2, we get:

⇒ a + 14d - (a + 3d) = 66 - 22

⇒ a + 14 d - a - 3d = 44

⇒ 11d = 44

⇒ d = $\frac{44}{11}$

⇒ d = 4.

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

⇒ a + 3(4) = 22

⇒ a + 12 = 22

⇒ a = 22 - 12

⇒ a = 10.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

⇒ S₈ = $\frac{8}{2}$ [2(10) + (8 - 1)4]

= 4[20 + (7)4]

= 4[20 + 28]

= 4 × (48)

= 192.

Hence, a = 10, d = 4, S₈ = 192.

Let a be the first term and d be the common difference.

We know that,

∴ a_n = a + (n - 1)d

Given,

The 4th term of an A.P. is 8.

⇒ a₄ = a + (4 - 1)d

⇒ 8 = a + 3d

⇒ a + 3d = 8 ....(1)

Given,

The 6th term of an A.P. is 14.

⇒ a₆ = a + (6 - 1)d

⇒ 14 = a + 5d

⇒ a + 5d = 14 ....(2)

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

⇒ a + 5d - (a + 3d) = 14 - 8

⇒ a + 5d - a - 3d = 6

⇒ 2d = 6

⇒ d = $\frac{6}{2}$

⇒ d = 3.

Substituting value of d in equation (1):

⇒ a + 3(3) = 8

⇒ a + 9 = 8

⇒ a = 8 - 9

⇒ a = -1.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

Now we have,

a = -1

d = 3

n = 20

⇒ S₂₀ = $\frac{20}{2}$ [2(-1) + (20 - 1)3]

= 10 [-2 + (19)3]

= 10 [-2 + 57]

= 10 × (55)

= 550.

Hence, a = -1, d = 3 and S₂₀ = 550.

Let the three consecutive terms of the A.P. be :

a - d, a, a + d

Given,

The sum of the first three terms is 42.

⇒ a - d + a + a + d = 42

⇒ 3a = 42

⇒ a = $\frac{42}{3}$

⇒ a = 14.

Given,

The product of the first term and the third term is 52:

⇒ (a - d)(a + d) = 52

⇒ (14 - d)(14 + d) = 52

⇒ (14)² - d² = 52

⇒ 196 - d² = 52

⇒ 196 - 52 = d²

⇒ 144 = d²

⇒ d² = 144

⇒ d = $\sqrt{144}$

⇒ d = 12 or -12

Case 1: a = 14 and d = 12

First term = a - d = 14 - 12 = 2

Case 2: a = 14 and d = -12

First term = a - d = 14 - (-12) = 26

Hence, first term = 2 and common difference = 12 or first term = 26 and common difference = -12.

Let a be the first term and d be the common difference.

We know that,

∴ a_n = a + (n - 1)d

Given,

6th term is equal to four times the first term.

⇒ a₆ = 4a₁

⇒ a + (6 - 1)d = 4a

⇒ 5d = 4a - a

⇒ 5d = 3a .........(1)

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

Given,

Sum of first six terms is 75.

⇒ S₆ = 75

⇒ $\frac{6}{2}$ [2a + (6 - 1)d] = 75

⇒ 3[2a + 5d] = 75

⇒ 2a + 5d = $\frac{75}{3}$

⇒ 2a + 5d = 25 ....(2)

Substituting value of 5d from equation (1) in (2), we get:

⇒ 2a + 3a = 25

⇒ 5a = 25

⇒ a = $\frac{25}{5}$

⇒ a = 5.

Substitute a = 5 into Equation 1, we get :

⇒ 5d = 3a

⇒ 5d = 3(5)

⇒ 5d = 15

⇒ d = $\frac{15}{5}$

⇒ d = 3.

Hence, a = 5 and d = 3.

Let a be the first term and d be the common difference.

We know that,

∴ a_n = a + (n - 1)d

Given,

The 5th term of an A.P. is 4.

⇒ a + (5 - 1)d = 4

⇒ a + 4d = 4 ....(1)

Given,

The 9th term of an A.P. is -12.

⇒ a + (9 - 1)d = -12

⇒ a + 8d = -12 ....(2)

Subtracting Equation (1) from Equation (2):

⇒ a + 8d - (a + 4d) = -12 - 4

⇒ a + 8d - a - 4d = -16

⇒ 4d = -16

⇒ d = $\frac{-16}{4}$

⇒ d = -4.

Substituting d = -4 in Equation (1) :

⇒ a + 4(-4) = 4

⇒ a - 16 = 4

⇒ a = 4 + 16

⇒ a = 20.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

Now we have,

a = 20

d = -4

n = 16

⇒ S₁₆ = $\frac{16}{2}$ [2(20) + (16 - 1)-4]

= 8[40 + (15)-4]

= 8[40 - 60]

= 8.(-20)

= -160.

Hence, a = 20, d = -4 and S₁₆ = -160.

The given A.P. is 15, 30, 45, 60,......

a = 15

d = 30 - 15 = 15

a_n = 300

We know that,

∴ a_n = a + (n - 1)d

⇒ 300 = 15 + (n - 1)15

⇒ 300 - 15 = (n - 1) 15

⇒ 285 = (n - 1)15

⇒ $\frac{285}{15}$ = n - 1

⇒ 19 = n - 1

⇒ n = 19 + 1

⇒ n = 20

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$(a + l)

Now we have,

a = 15

n = 20

l = 300

⇒ S₂₀ = $\frac{20}{2}$ (15 + 300)

= 10 (315)

= 3150

Hence, S₂₀ = 3150.

Given,

First term (a) = 164

Common difference (d) = 160 - 164 = -4

(i) Let n^th term be zero.

⇒ a_n = 0

⇒ a + (n - 1)d = 0

⇒ 164 + (n - 1)(-4) = 0

⇒ 164 - 4n + 4 = 0

⇒ 168 - 4n = 0

⇒ 4n = 168

⇒ n = $\frac{168}{4}$

⇒ n = 42.

Hence, 42nd term is equal to 0.

(ii) By formula,

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

Substituting values we get :

$$\begin{aligned}S_{20} = \frac{20}{2}\Big[2 \times (164) + (20 - 1) \times (-4)\Big] \\[1em] \Rightarrow S_{20} = 10[328 + (19) \times (-4)] \\[1em] \Rightarrow S_{20} = 10[328 - 76] \\[1em] \Rightarrow S_{20} = 10 \times 252 \\[1em] \Rightarrow S_{20} = 2520.\end{aligned}$$

Hence, sum of first 20 terms = 2520.

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 = $\frac{3}{4}$.

Given,

Each prize is ₹ 200 less than the preceding prize.

Prizes : a, a - 200, a - 400, a - 600.

The above is an A.P., wth first term = a and common difference = ₹ -200.

Total prize money = ₹ 2,800.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 2800 = \frac{4}{2} \times [2 \times a + (4 - 1) \times -200] \\[1em] \Rightarrow 2800 = 2 \times [2a - 600] \\[1em] \Rightarrow 2800 = 4a - 1200 \\[1em] \Rightarrow 4a = 2800 + 1200 \\[1em] \Rightarrow 4a = 4000 \\[1em] \Rightarrow a = \frac{4000}{4} = ₹ 1,000.\end{aligned}$$

a - ₹ 200 = ₹ 1,000 - ₹ 200 = ₹ 800

a - ₹ 400 = ₹ 1,000 - ₹ 400 = ₹ 600

a - ₹ 600 = ₹ 1,000 - ₹ 600 = ₹ 400.

Hence, 1st Prize = ₹ 1,000, 2nd Prize = ₹ 800, 3rd Prize = ₹ 600, 4th Prize = ₹ 400.

(i) Given,

The production of TV sets in a factory increases uniformly by a fixed number every year. So the production is in A.P.

Let the TV sets produced in first year be a and difference in production of each year be d.

We know that,

∴ a_n = a + (n - 1)d

Given,

The sets produced in 6 years is 8000.

⇒ a + (6 - 1)d = 8000

⇒ a + 5d = 8000 ....(1)

The sets produced in the 9th year is 11300.

⇒ a + (9 - 1)d = 11300

⇒ a + 8d = 11300 ....(2)

Subtract Equation (1) from Equation (2), we get:

⇒ a + 8d - (a + 5d) = 11300 - 8000

⇒ 3d = 3300

⇒ d = $\frac{3300}{3}$

⇒ d = 1100.

Substitute d = 1100 into Equation 1 :

⇒ a + 5(1100) = 8000

⇒ a + 5500 = 8000

⇒ a + 5500 = 8000

⇒ a = 8000 - 5500

⇒ a = 2500.

Hence, the production in the first year is 2500 sets.

(ii) Solving,

⇒ a₈ = a + 7d

= 2500 + 7(1100)

= 2500 + 7700

= 10200.

Hence, production in 8th year = 10200.

(iii) Total production in 6 years :

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

⇒ S₆ = $\frac{6}{2}$ [2(2500) + (6 - 1)1100]

= 3[5000 + (5)1100]

= 3[5000 + 5500]

= 3(10500)

= 31500.

Hence, total production in 6 years = 31500.

The number of logs in each row forms an Arithmetic Progression (A.P.):

20, 19, 18,......

Total sum of logs = 200.

S_n = 200

Common difference (d) = 19 - 20 = -1

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

⇒ 200 = $\frac{n}{2}$ [2(20) + (n - 1)(-1)]

⇒ 200 × 2 = n[40 - n + 1]

⇒ 400 = n(41 - n)

⇒ 400 = 41n - n²

⇒ n² - 41n + 400 = 0

⇒ n² - 16n - 25n + 400 = 0

⇒ n(n - 16) - 25(n - 16) = 0

⇒ (n - 25)(n - 16) = 0

⇒ (n - 25) = 0 or (n - 16) = 0

⇒ n = 16 or n = 25.

Case 1 :

If n = 25,

logs in the 25th row a₂₅:

⇒ a₂₅ = a + (25 - 1)d

= 20 + (24)(-1)

= 20 - 24

= -4 (the number of logs cannot be negative).

Thus, n ≠ 25.

If n = 16

logs in the 16th row a₁₆:

⇒ a₁₆ = a + (16 - 1)d

= 20 + (15)(-1)

= 20 - 15

= 5.

Hence, no. of rows = 16 and the top row has 5 logs.

Given,

The number of rose plants in each row forms an Arithmetic Progression (A.P.):

43, 41, 39,......., 11.

a = 43

l = 11

d = 43 - 41 = -2

We know that,

∴ a_n = a + (n - 1)d

⇒ 11 = 43 + (n - 1)(-2)

⇒ 11 - 43 = (n - 1)(-2)

⇒ -32 = (n - 1)(-2)

⇒ $\frac{-32}{-2}$ = (n - 1)

⇒ n - 1 = 16

⇒ n = 16 + 1

⇒ n = 17

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$(a + l)

Total number of rose plants:

⇒ S₁₇ = $\frac{17}{2}$(43 + 11)

= $\frac{17}{2}$(43 + 11)

= 15 × $\frac{54}{2}$

= 17 × 27

= 459.

Hence, number of rows = 17, total number of rose plants in the bed is 459.

Let the man saved ₹ a in first month.

The amount the man saved each month forms an Arithmetic Progression (A.P.) where :

Total sum saved S_n = ₹ 33,000.

Number of months (n) = 10

He saves ₹ 100 more than he did in the preceding month.

Thus, d = 100

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$[2a + (n - 1)d]

⇒ 33000 = $\frac{10}{2}$ [2a + (10 - 1)100]

⇒ 33000 = 5[2a + (9)100]

⇒ $\frac{33000}{5}$ = [2a + 900]

⇒ 6600 = [2a + 900]

⇒ 6600 - 900 = 2a

⇒ 2a = 5700

⇒ a = $\frac{5700}{2}$

⇒ a = ₹ 2,850.

Hence, the amount man saved in first month is ₹ 2,850.

In the A.P. :

$\frac{1}{a}, \Big(\frac{1}{a} + 1\Big), \Big(\frac{1}{a} + 2\Big), ......$

First term (x) = $\frac{1}{a}$

Common difference (d) = $\frac{1}{a} + 1 - \frac{1}{a}$ = 1.

x₇ = x + (7 - 1)d

= $\frac{1}{a} + 6 \times 1$

= $\Big(\frac{1}{a} + 6\Big)$.

Hence, Option 1 is the correct option.

Since, we need to calculate term from end.

So, first term will be equal to last term and common difference will be negative of the original difference.

∴ T_end = l + (n - 1)(-d)

= l - (n - 1)d.

Hence, option 2 is the correct option.

The sequence of natural numbers 1, 2, 3,....., n.

a = 1

l = n

no. of terms = n

Sum of n terms of an A.P. is given by,

∴ S_n = $\frac{n}{2}$ (a + l)

= $\frac{n}{2}$ (1 + n)

= $\frac{n(n + 1)}{2}$

Hence, option 2 is the correct option.

Sequence : 1, 2, 3, ......, 50.

First term (a) = 1

Common difference (d) = 2 - 1 = 1

By formula,

Sum of n terms = $\frac{n}{2}[2a + (n - 1)d]$

$$\begin{aligned}= \frac{50}{2} \times [2 \times 1 + (50 - 1) \times 1] \\[1em] = 25 \times [2 + 49] \\[1em] = 25 \times 51 \\[1em] = 1275.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

nth term of A.P. :

∴ T_n = 2n + 5

⇒ T₁₀ = 2(10) + 5

= 20 + 5

= 25.

Hence, option 3 is the correct option.

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

Given,

a = 7, d = 3

T_n = 7 + (n - 1)3

= 7 + 3n - 3

= 3n + 4.

Hence, option 3 is the correct option.

The given Arithmetic Progression (A.P.) is

-1, 3, 7, 11,.....

a = -1

d = 3 - (-1) = 4

n = 24

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ T₂₄ = -1 + (24 - 1)(4)

= -1 + (23)4

= -1 + 92

= 91.

Hence, option 3 is the correct option.

The given Arithmetic Progression (A.P.) is

70, 75, 80, 85, ....

a = 70

d = 75 - 70 = 5

n = 10

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ T₁₀ = 70 + (10 - 1)5

= 70 + (9)5

= 70 + 45

= 115.

Hence, option 3 is the correct option.

Given,

A.P. 6, 13, 20, …, 216

n = 31

Middle term of A.P. = $\frac{n + 1}{2}$

= $\frac{31 + 1}{2}$

= $\frac{32}{2}$

= 16th term.

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

Now we have,

a = 6

d = 13 - 6 = 7

n = 16

⇒ T₁₆ = 6 + (16 - 1)(7)

= 6 + (15)7

= 6 + 105

= 111.

Hence, option 4 is the correct option.

Given,

A.P. : 7, 13, 19, 25,.......

a = 7

d = 13 - 7 = 6

Let nth term be 241.

T_n = 241

⇒ a + (n - 1)d = 241

⇒ 241 = 7 + (n - 1)6

⇒ 241 - 7 = (n - 1)6

⇒ 234 = (n - 1)6

⇒ $\frac{234}{6}$ = n - 1

⇒ n - 1 = 39

⇒ n = 39 + 1

⇒ n = 40.

Hence, option 1 is the correct option.

Given,

A.P. : 11, 8, 5, 2, .......

a = 11

d = 8 - 11 = -3

Let nth term of the A.P. be -148.

T_n = -148

⇒ a + (n - 1)d = -148

⇒ -148 = 11 + (n - 1)(-3)

⇒ -148 - 11 = (n - 1)(-3)

⇒ -159 = (n - 1)(-3)

⇒ $\frac{-159}{-3}$ = n - 1

⇒ n - 1 = 53

⇒ n = 53 + 1

⇒ n = 54.

Hence, option 2 is the correct option.

The three-digit numbers divisible by 7 form an Arithmetic Progression (A.P.) :

105, 112, 119, ......,994.

a = 105

l = 994

d = 7

Let no. of terms be n.

∴ T_n = a + (n - 1)d

⇒ 994 = 105 + (n - 1)7

⇒ 994 - 105 = (n - 1)7

⇒ 889 = (n - 1)7

⇒ $\frac{889}{7}$ = n - 1

⇒ 127 = n - 1

⇒ n = 127 + 1

⇒ n = 128.

Hence, option 1 is the correct option.

The numbers divisible by 4 lying between 20 and 200 form an Arithmetic Progression (A.P.).

24, 28, ....., 196.

a = 24

l = 196

d = 4

Let no. of terms be n.

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ 196 = 24 + (n - 1)4

⇒ 196 - 24 = (n - 1)4

⇒ 172 = (n - 1)4

⇒ $\frac{172}{4}$ = n - 1

⇒ 43 = n - 1

⇒ n = 43 + 1

⇒ n = 44.

Hence, option 2 is the correct option.

Given,

A.P. : $\frac{1}{k}, \frac{1 - k}{k}, \frac{1 - 2k}{k},$ .........

a = $\frac{1}{k}$

$$\begin{aligned}\Rightarrow d = \frac{1 - k}{k} - \frac{1}{k} \\[1em] \Rightarrow d = \frac{1 - k - 1}{k} \\[1em] \Rightarrow d = \frac{-k}{k} \\[1em] \Rightarrow d = -1.\end{aligned}$$

Hence, option 4 is the correct option.

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

T_n = 3n + 2

T₁ = 3(1) + 2 = 5

T₂ = 3(2) + 2 = 8

T₃ = 3(3) + 2 = 11

Sum = 5 + 8 + 11 = 24.

Hence, option 2 is the correct option.

Given,

A.P. : 5, 12, 19, .......

a = 5

d = 12 - 5 = 7

n = 60

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ T₆₀ = 5 + (60 - 1)7

= 5 + (59) × 7

= 5 + 413

= 418.

Hence, option 4 is the correct option.

In the above A.P.,

$$\begin{aligned}\Rightarrow d = \frac{(1 − 3p)}{3} - \frac{1}{3} \\[1em] \Rightarrow d = \frac{(1 − 3p) - 1}{3} \\[1em] \Rightarrow d = \frac{-3p}{3} \\[1em] \Rightarrow d = -p.\end{aligned}$$

Hence, option 2 is the correct option.

In the above A.P. next terms is the 5th term.

a = $\sqrt{3}$

n = 5

$$\begin{aligned}\Rightarrow d = \sqrt{12} - \sqrt{3} \\[1em] \Rightarrow d = \sqrt{4 \times 3} - \sqrt3 \\[1em] \Rightarrow d = 2\sqrt{3} - \sqrt3 \\[1em] \Rightarrow d = \sqrt3.\end{aligned}$$

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

$$\begin{aligned}\Rightarrow T_5 = \sqrt3 + (5 - 1)\sqrt3 \\[1em] = \sqrt3 + 4\sqrt3 \\[1em] = 5\sqrt3 \\[1em] = \sqrt{5^2 \times 3} \\[1em] = \sqrt{75}.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

First term = p

Common difference = q

n = 10

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ T₁₀ = p + (10 - 1)q

= p + 9q.

Hence, option 4 is the correct option.

Given,

a = $\frac{1}{m}$

$$\begin{aligned}\Rightarrow d = \frac{(1 + m)}{m} - \frac {1}{m} \\[1em] \Rightarrow d = \frac{(1 + m) - 1}{m} \\[1em] \Rightarrow d = \frac{m}{m} \\[1em] \Rightarrow d = 1.\end{aligned}$$

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

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

Hence, option 1 is the correct option.

Given,

2p + 1, 13, 5p − 3

In an A.P., the middle term is the average of the first and third terms.

$$\begin{aligned}\Rightarrow 13 = \frac{(2p + 1) + (5p - 3)}{2} \\[1em] \Rightarrow 13 = \frac{7p - 2}{2} \\[1em] \Rightarrow 13 \times 2 = 7p - 2 \\[1em] \Rightarrow 26 = 7p - 2 \\[1em] \Rightarrow 26 + 2 = 7p \\[1em] \Rightarrow 28 = 7p \\[1em] \Rightarrow 7p = 28 \\[1em] \Rightarrow p = \frac{28}{7} \\[1em] \Rightarrow p = 4.\end{aligned}$$

Hence, option 4 is the correct option.

We are given three consecutive terms of an A.P.:

2k + 1, 3k + 3, 5k − 1

In an arithmetic progression, the difference between consecutive terms is the same.

⇒ (3k + 3) - (2k + 1) = (5k - 1) - (3k + 3)

⇒ 3k + 3 - 2k - 1 = 5k - 1 - 3k - 3

⇒ k + 2 = 2k - 4

⇒ 2 + 4 = 2k - k

⇒ k = 6.

Hence, option 4 is the correct option.

Given,

S_n = 5n² + 3n

We know that,

nth term of A.P. :

∴ T_n = S_n - S_{n - 1}

= 5n² + 3n - [5(n - 1)² + 3(n - 1)]

= 5n² + 3n - [5(n² - 2n + 1) + 3n - 3]

= 5n² + 3n - [5n² - 10n + 5 + 3n - 3]

= 5n² + 3n - 5n² + 10n - 5 - 3n + 3

= 10n - 2

d = a_n - a_{n - 1}

= 10n - 2 - [10(n - 1) - 2]

= 10n - 2 - [10n - 10 - 2]

= 10n - 2 - 10n + 10 + 2

= 10.

Hence, option 2 is the correct option.

Given,

a = 1

l = 11

Let no. of terms be n.

S_n = 36

We know that,

⇒ S_n = $\frac{n}{2}$ (a + l)

⇒ 36 = $\frac{n}{2}$ (1 + 11)

⇒ 36 × 2 = n(12)

⇒ n = $\frac{36 \times 2}{12}$

⇒ n = 6.

Hence, option 1 is the correct option.

Sequence :

6, 12, 18, ......., upto 40th term.

The above sequence is an A.P. with,

a = 6

d = 6

n = 40

We know that,

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

⇒ S₄₀ = $\frac{40}{2}$ [2(6) + (40 - 1)6]

= 20[12 + (39)6]

= 20(12 + 234)

= 20 × (246)

= 4920.

Hence, option 3 is the correct option.

1, 3, 5, 7, ......, 30th term.

The odd natural numbers form an Arithmetic Progression (A.P.) :

a = 1

d = 3 - 1 = 2

n = 30

We know that,

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

⇒ S₃₀ = $\frac{30}{2}$ [2(1) + (30 - 1)2]

= 15[2 + (29)2]

= 15[2 + 58]

= 15 × (60)

= 900.

Hence, option 2 is the correct option.

Given,

a₇ = 4

d = -4

n = 7

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₇ = a + (7 - 1)(-4)

⇒ 4 = a + (6)(-4)

⇒ 4 = a - 24

⇒ 24 + 4 = a

⇒ a = 28.

Hence, option 2 is the correct option.

We know that,

S_n = 4n² + 2n

T_n = S_n - S_{n - 1}

= 4n² + 2n - [4(n - 1)² + 2(n - 1)]

= 4n² + 2n - [4(n² - 2n + 1) + 2n - 2]

= 4n² + 2n - [4n² - 8n + 4 + 2n - 2]

= 4n² + 2n -[4n² - 6n + 2]

= 4n² + 2n - 4n² + 6n - 2

= 8n - 2.

Hence, option 2 is the correct option.

Given,

a = 7

⇒ a₁₃ = 35

⇒ a + (13 - 1)d = 35

⇒ 7 + 12d = 35

⇒ 12d = 35 - 7

⇒ 12d = 28

⇒ d = $\frac{28}{12}$

⇒ d = $\frac{7}{3}$.

Hence, option 4 is the correct option.

We know that,

S_n = $\frac{n}{2}$(a + l)

Given,

n = 9

l = 28

S_n = 144

⇒ S₉ = $\frac{9}{2}$(a + 28)

⇒ 144 = $\frac{9}{2}$(a + 28)

⇒ $\frac{144 \times 2}{9}$ = (a + 28)

⇒ 32 = (a + 28)

⇒ 32 - 28 = a

⇒ a = 4.

Hence, option 3 is the correct option.

Given,

a₅ = -3

d = -4

n = 10

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₅ = a + (5 - 1)(-4)

⇒ -3 = a + (4)(-4)

⇒ -3 = a - 16

⇒ a = 16 - 3

⇒ a = 13.

We know that,

⇒ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

⇒ S₁₀ = $\frac{10}{2}$ [2(13) + (10 - 1)(-4)]

= 5[26 + (9)(-4)]

= 5[26 - 36]

= 5(-10)

= -50.

Hence, option 1 is the correct option.

We know that,

a_n = a + (n - 1)d

The 7th term of an A.P. is −1.

a + 6d = -1 ......(1)

The 16th term of an A.P. is 17.

a + 15d = 17 ......(2)

Subtract Equation (1) from Equation (2) :

⇒ a + 15d - (a + 6d) = 17 - (-1)

⇒ a + 15d - a - 6d = 18

⇒ 9d = 18

⇒ d = $\frac{18}{9}$

⇒ d = 2.

Substituting d = 2 into Equation (1), we get :

⇒ a + 6(2) = -1

⇒ a + 12 = -1

⇒ a = -1 - 12

⇒ a = -13.

Now,

⇒ a_n = a + (n - 1)d

⇒ a_n = -13 + (n - 1)2

= -13 + 2n - 2

= 2n - 15.

Hence, option 4 is the correct option.

Given,

⇒ S_p = ap² + bp

⇒ T_p = S_p - S_{p - 1}

= ap² + bp - [a(p - 1)² + b(p - 1)]

= ap² + bp - [a(p² - 2p + 1) + bp - b]

= ap² + bp - [ap² - 2ap + a + bp - b]

= ap² + bp - ap² + 2ap - a - bp + b

= 2ap - a + b.

d = T_p - T_{p - 1}

= 2ap - a + b - [2a(p - 1) - a + b]

= 2ap - a + b - [2ap - 2a - a + b]

= 2ap - a + b - 2ap + 2a + a - b

= 2a.

Hence, option 4 is the correct option.

Given,

First term = a

nth term = b

We know that,

a_n = a + (n - 1)d

⇒ b = a + (n - 1)d

⇒ b - a = (n - 1)d

⇒ d = $\frac{b - a}{n - 1}$.

Hence, option 2 is the correct option.

Given,

The first three terms of an A.P. are 3y − 1, 3y + 5 and 5y + 1.

The difference between consecutive terms must be equal.

⇒ 3y + 5 - (3y - 1) = 5y + 1 - (3y + 5)

⇒ 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5

⇒ 6 = 2y - 4

⇒ 2y = 6 + 4

⇒ 2y = 10

⇒ y = $\frac{10}{2}$

⇒ y = 5.

Hence, option 2 is the correct option.

Given,

a = 7

d = 10 - 7 = 3

a_n = 184

We know that,

⇒ a_n = a + (n - 1)d

⇒ a_n = 7 + (n - 1)(3)

⇒ 184 = 7 + (n - 1)(3)

⇒ 184 - 7 = (n - 1)(3)

⇒ 177 = (n - 1)(3)

⇒ $\frac{177}{3}$ = (n - 1)

⇒ n - 1 = 59

⇒ n = 59 + 1

⇒ n = 60.

The 8th term from the end is the (n - 8 + 1)th term from the beginning.

= 60 - 8 + 1

= 53.

⇒ a₅₃ = 7 + (53 - 1)3

= 7 + (52)(3)

= 7 + 156

= 163.

Hence, option 3 is the correct option.

a = 3

n = 8

S_n = 192

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

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

⇒ 192 = 4[6 + 7d]

⇒ $\frac{192}{4}$ = [6 + 7d]

⇒ 48 = [6 + 7d]

⇒ 48 - 6 = 7d

⇒ 42 = 7d

⇒ d = $\frac{42}{7}$

⇒ d = 6.

Hence, option 3 is the correct option.

Given,

Sum of n terms of an arithmetic progression S_n = n² - n.

S₁ = 1² - 1 = 0,

S₂ = 2² - 2 = 4 - 2 = 2,

S₃ = 3² - 3 = 9 - 3 = 6.

Sum upto first term = First term = 0.

Given, sum upto 2 terms = 2 and first term = 0, second term = 2.

Sum upto third term = 6

∴ First term + Second term + Third term = 6

⇒ 0 + 2 + Third term = 6

⇒ Third term = 6 - 2 = 4.

Hence, Option 2 is the correct option.

A.P. : -1, 5, 11, ......

Given,

a = -1

d = 5 - (-1) = 6

We know that,

S_n = $\frac{n}{2}$ [2a + (n − 1)d]

⇒ S_n = $\frac{n}{2}$ [2(-1) + (n − 1)6]

= $\frac{n}{2}$ [-2 + (6n − 6)]

= $\frac{n}{2}$ (6n − 8)

= $\frac{n}{2} \times$ 2(3n − 4)

= n(3n - 4)

= 3n² - 4n.

∴ Assertion (A) is true.

The standard and correct formula for the sum of the first n terms of an A.P.

S_n = $\frac{n}{2}$[2a + (n − 1)d]

∴ Reason (R) is true.

Both A and R are true, and R is the correct explanation of A.

Hence, option 1 is the correct option.

Given,

a = 17

a_n = -40

d = 14 - 17 = -3

We know that,

⇒ a_n = a + (n - 1)d

⇒ -40 = 17 + (n - 1)(-3)

⇒ -40 - 17 = (n - 1)(-3)

⇒ -57 = (n - 1)(-3)

⇒ $\frac{-57}{-3}$ = (n - 1)

⇒ 19 = n - 1

⇒ n = 19 + 1

⇒ n = 20.

The A.P. has 20 terms.

The 10th term from the end is the (n - 10 + 1)th term from the beginning

= 20 - 10 + 1 = 11th term from beginning.

⇒ a_n = a + (n - 1)d

⇒ a₁₁ = 17 + (11 - 1)(-3)

= 17 + 10(-3)

= 17 - 30

= -13.

Assertion (A) is false.

The standard and correct formula for finding the nth term of an Arithmetic Progression.

a_n = a + (n - 1)d

Reason (R) is true.

A is false, R is true

Hence, option 4 is the correct option.

Given,

a₂₂ = 149

n = 22

d = 7.

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₂₂ = a + (22 - 1)7

⇒ 149 = a + (21)7

⇒ 149 = a + 147

⇒ 149 - 147 = a

⇒ a = 2.

We know that,

S_n = $\frac{n}{2}$ (a + l)

⇒ S₂₂ = $\frac{22}{2}$ (2 + 149)

= 11 × (151)

= 1661.

Assertion (A) is true.

The standard and correct formula for the sum of the first n terms of an A.P.

S_n = $\frac{n}{2}$ [2a + (n − 1)d]

Reason (R) is false.

A is true, R is false

Hence, option 3 is the correct option.

We know that,

T_n = S_n - S_{n - 1}

Given,

S_n = 2n² - n

S_{n - 1} = 2(n - 1)² - (n - 1)

= 2(n² - 2n + 1) - n + 1

= 2n² - 4n + 2 - n + 1

= 2n² - 5n + 3.

T_n = 2n² - n - (2n² - 5n + 3)

= 2n² - n - 2n² + 5n - 3

= 4n - 3.

Assertion (A) is true.

The nth term (t_n) of an A.P. from the end is l - (n - 1)d.

Reason (R) is true.

Both A and R are true, but R is not the correct explanation of A.

Hence, option 2 is the correct option.

(a) A.P. = 21, 42, 63, .........

The above sequence is an A.P. with first term (a) = 21 and common difference (d) = 21.

By formula,

T_n = a + (n - 1)d

= 21 + (n - 1)21

= 21 + 21n - 21

= 21n.

Hence, the n^th term = 21n.

(b) A.P. = 21, 42, 63, 84, 105, ........

Hence, there are four two digit numbers i.e. 21, 42, 63, 84 in the A.P.

(c) By formula,

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

$$\begin{aligned}\text{Sum of first 10 terms of A.P.} = \frac{10}{2}(a + a_{10}) \\[1em] = 5[a + a + (n - 1)d] \\[1em] = 5[2a + (n - 1)d] \\[1em] = 5[2 \times 21 + (10 - 1) \times 21] \\[1em] = 5[42 + 9 \times 21] \\[1em] = 5[42 + 189] \\[1em] = 5 \times 231 \\[1em] = 1155.\end{aligned}$$

Hence, sum of first 10 terms of the A.P. = 1155.

(a) Let no. of terms be n.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 65 = \frac{n}{2}[2 \times 20 + (n - 1) \times (-3)] \\[1em] \Rightarrow 65 = \frac{n}{2}[40 - 3n + 3] \\[1em] \Rightarrow 65 = \frac{n}{2}[43 - 3n] \\[1em] \Rightarrow 65 \times 2 = n(43 - 3n) \\[1em] \Rightarrow 130 = 43n - 3n^2 \\[1em] \Rightarrow 3n^2 - 43n + 130 = 0 \\[1em] \Rightarrow 3n^2 - 30n - 13n + 130 = 0 \\[1em] \Rightarrow 3n(n - 10) - 13(n - 10) = 0 \\[1em] \Rightarrow (3n - 13)(n - 10) = 0 \\[1em] \Rightarrow 3n - 13 = 0 \text{ or } n - 10 = 0 \\[1em] \Rightarrow 3n = 13 \text{ or } n = 10 \\[1em] \Rightarrow n = \frac{13}{3} \text{ or } n = 10.\end{aligned}$$

Since, no. of terms cannot be in fraction.

∴ n = 10.

Hence, no. of terms = 10.

(b) By formula,

Last term (l) = a + (n - 1)d

= 20 + (10 - 1) × (-3)

= 20 + 9 × (-3)

= 20 - 27

= -7.

Hence, last term = -7.

(a) Difference between common terms : 16 - 9 = 9 - 2 = 7.

Since, difference between common terms are equal.

Hence, the given sequence is an A.P.

(b) By formula,

a_n = a + (n - 1)d

a₂₀ = 2 + (20 - 1) × 7

= 2 + 19 × 7

= 2 + 133

= 135.

Hence, 20th term of the sequence = 135.

(c) By formula,

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

Substituting values we get :

Sum upto 22 terms :

$$\begin{aligned}S_{22} = \frac{22}{2}[2 \times 2 + (22 - 1) \times 7] \\[1em] = 11 \times [4 + 21 \times 7] \\[1em] = 11 \times [4 + 147] \\[1em] = 11 \times 151 \\[1em] = 1661.\end{aligned}$$

Sum upto 25 terms :

$$\begin{aligned}S_{25} = \frac{25}{2}[2 \times 2 + (25 - 1) \times 7] \\[1em] = \frac{25}{2} \times [4 + 24 \times 7] \\[1em] = \frac{25}{2} \times [4 + 168] \\[1em] = \frac{25}{2} \times 172 \\[1em] = 25 \times 86 \\[1em] = 2150.\end{aligned}$$

S₂₅ - S₂₂ = 2150 - 1661 = 489.

Hence, difference between the sum of its first 22 and 25 terms = 489.

(d) Let n^th term be 102.

⇒ a_n = a + (n - 1)d

⇒ 102 = 2 + 7(n - 1)

⇒ 102 = 2 + 7n - 7

⇒ 102 = 7n - 5

⇒ 102 + 5 = 7n

⇒ 7n = 107

⇒ n = $\frac{107}{7} = 15\frac{2}{7}$.

Since, n cannot be in fraction.

Hence, 102 is not the term of the sequence.

(e) If k is added to each term.

Sequence : 2 + k, 9 + k, 16 + k,.............

The common difference between terms is still equal to 7.

Hence, sequence is in A.P.

1. 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ....

The radius are in A.P. with

a = 0.5 cm

d = 1.0 - 0.5 = 0.5 cm

n = 9

We know that,

⇒ a_n = a + (n - 1)d

⇒ r₉ = 0.5 + (9 - 1)0.5

= 0.5 + (8)0.5

= 0.5 + 4

= 4.5 cm

Hence, option (b) is the correct option.

2. Calculating the radius of 15th semicircle.

We know that,

⇒ a_n = a + (n - 1)d

⇒ r₁₅ = 0.5 + (15 - 1)0.5

= 0.5 + 14(0.5)

= 0.5 + 7

= 7.5 cm

By formula,

Length of a semicircle = πr

= 7.5π cm

Hence, option (c) is the correct option.

3. We know that,

Length of a semicircle = πr

The lengths are: 0.5π cm, 1.0π cm, 1.5π cm, 2.0π cm,.... in an A.P. with,

a = 0.5π cm

d = 1.0π cm - 0.5π cm = 0.5π

The difference of lengths of the 19th and 12th semicircles is:

L₁₉ - L₁₂ = πr₁₉ - πr₁₂

= π(r₁₉ - r₁₂)

We know that,

a_n = a + (n - 1)d

⇒ r₁₉ = 0.5π + (19 - 1)0.5π

= 0.5π + 18(0.5π)

= 0.5π + 9π

= 9.5π cm.

⇒ r₁₂ = 0.5π + (12 - 1)0.5π

= 0.5π + 11(0.5π)

= 0.5π + 5.5π

= 6π cm.

L₁₉ - L₁₂ = (9.5π - 6π)

= 3.5π cm.

Hence, option (a) is the correct option.

4. The lengths are: 0.5π cm, 1.0π cm, 1.5π cm, 2.0π cm,....in an A.P.

a = 0.5π cm

d = 1.0π cm - 0.5π cm = 0.5π

The total length is the sum of the first 7 lengths S₇.

We know that,

⇒ S_n = $\frac{n}{2}$ [2a + (n - 1)d]

⇒ S₇ = $\frac{7}{2}$ [2(0.5π) + (7 - 1)(0.5π)]

= 3.5[1π + 6(0.5π)]

= 3.5[1π + 3π]

= 3.5(4π)

= 14π cm.

Total length = 14π cm

= $14 \times \frac{22}{7}$

= 44 cm.

Hence, option (c) is the correct option.

5. 0.5π cm, 1.0π cm, 1.5π cm, 2.0π cm, ....

d = 1.0π + 0.5π = 0.5 π cm.

Common difference of the A.P. of lengths = 0.5π cm.

Hence, option (a) is the correct option.

1. Since, production of TV sets in a factory increases uniformly by a fixed number every year. The production in each year forms an A.P.

Let production of the TV sets during the first year be a and the common difference in production each year be d.

We know that,

a_n = a + (n - 1)d

Given,

Production in the 6th year = 16000.

⇒ a₆ = a + (6 - 1)d

⇒ 16000 = a + 5d

⇒ a + 5d = 16000 ....(1)

Given,

Production in the 9th year = 22600.

⇒ a₉ = a + (9 - 1)d

⇒ 22600 = a + 8d

⇒ a + 8d = 22600 ....(2)

Subtracting Eq. (1) from Eq. (2), we get :

⇒ a + 8d - (a + 5d) = 22600 - 16000

⇒ a + 8d - a - 5d = 6600

⇒ 3d = 6600

⇒ d = $\frac{6600}{3}$

⇒ d = 2200.

Substituting d in equation (1), we get :

⇒ a + 5(2200) = 16000

⇒ a + 11000 = 16000

⇒ a = 16000 - 11000

⇒ a = 5000.

Hence, option (c) is the correct option.

2. From question (1),

The uniform increase is the common difference, d.

d = 2200.

Hence, option (d) is the correct option.

3. Given,

The production of TV sets in each year forms an A.P.: 5000, 7200, 9400, .....

a = 5000

n = 8

d = 2200

We know that,

a_n = a + (n - 1)d

⇒ a₈ = 5000 + (8 - 1)2200

= 5000 + 7(2200)

= 5000 + 15400

= 20400.

Production of the TV sets during the 8th year = 20400.

Hence, option (b) is the correct option.

4. Given,

The production of TV sets in each year forms an A.P.: 5000, 7200, 9400.....

a = 5000

l = 16000

n = 6

We know that,

S_n = $\frac{n}{2}$ (a + l)

⇒ S₆ = $\frac{6}{2}$ (5000 + 16000)

= 3 × (21000)

= 63000.

Total production of the TV sets during first 6 years = 63000

Hence, option (d) is the correct option.

5. From part 4 we have,

S₆ = 63000

The average production is the total production divided by the number of years:

Average = $\frac{S_6}{6}$

= $\frac{63000}{6}$

= 10500.

Hence, option (a) is the correct option.

1. Given,

Logs stacked in rows form an A.P.: 20, 19, 18.....

S_n = 200

a = 20

d = 19 - 20 = -1

We know that,

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

Let 200 logs be stacked in n rows.

⇒ 200 = $\frac{n}{2}$ [2(20) + (n - 1)(-1)]

⇒ 200 × 2 = n[40 - n + 1]

⇒ 400 = n[41 - n]

⇒ 400 = 41n - n²

⇒ n² - 41n + 400 = 0

⇒ n² - 16n - 25n + 400 = 0

⇒ n(n - 16) - 25(n - 16) = 0

⇒ (n - 25)(n - 16) = 0

⇒ (n - 25) = 0 or (n - 16) = 0

⇒ n = 25 or n = 16.

If n = 25, a₂₅ = a + 24d = 20 + 24(-1) = -4 (Impossible, the 25th row would have –4 logs).

If n = 16, a₁₆ = a + 15d = 20 + 15(-1) = 5 (Valid).

The number of rows is 16.

Hence, option (c) is the correct option.

2. Given,

Logs stacked in rows form an A.P.: 20, 19, 18, .....

a = 20

n = 16 (16 total rows of logs in the stack i.e top row is 16th.)

d = -1

We know that,

a_n = a + (n - 1)d

⇒ a₁₆ = 20 + (16 - 1)(-1)

= 20 + 15(-1)

= 20 - 15

= 5.

The number of logs in the top row is 5.

Hence, option (b) is the correct option.

3. Given,

Logs stacked in rows form an A.P.: 20, 19, 18.....

a = 20

Total rows = 16

n = 8, (A.P. starts from bottom only)

d = -1

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₈ = 20 + (8 - 1)(-1)

= 20 + 7(-1)

= 20 - 7

= 13.

The number of logs in the 8th row from the bottom is 13.

Hence, option (d) is the correct option.

4. Given,

Logs stacked in rows form an A.P.: 20, 19, 18, .....

a = 20

d = -1

n = 6

We know that,

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

⇒ S₆ = $\frac{6}{2}$ [2(20) + (6 - 1)(-1)]

= 3(40 - 5)

= 3 × (35)

= 105.

Total number of logs in the first six rows from the bottom is 105.

Hence, option (a) is the correct option.

5. The 5th row from the top is the (16 - 5 + 1) = 12th row from the bottom (a₁₂):

Logs stacked in rows form an A.P.: 20, 19, 18.....

We know that,

a_n = a + (n - 1)d

⇒ a₁₂ = 20 + (12 - 1)(-1)

= 20 - 11

= 9.

Hence, option (b) is the correct option.