15, 30, 60, 120 ...... are in G.P. (Geometric Progression).
(a) Find the nth term of this G.P. in terms of n.
(b) How many terms of the above G.P. will give the sum 945 ?
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Given,
G.P. : 15, 30, 60, 120 ......
First term (a) = 15
Common ratio (r) = $\frac{30}{15}$ = 2
(a) nth term of G.P. = arn - 1 = 15 x 2n - 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.