If the 6th term of a series in Geometric Progression (G.P.) is 32 and the 9th term is 256, find the:

(a) Let the first term of the Geometric Progression be a and the common ratio be r.

By formula,

T_n = arⁿ

Given,

The 6th term is 32.

⇒ ar^{6 - 1} = 32

⇒ ar⁵ = 32 .........(1)

The 9th term is 256.

⇒ ar^{9 - 1} = 256

⇒ ar⁸ = 256 .........(2)

Divide equation (2) by (1), we get:

$$\begin{aligned}\Rightarrow \frac{a \times r^8}{a \times r^5} = \frac{256}{32} \\[1em] \Rightarrow r^{8 - 5} = 8 \\[1em] \Rightarrow r^{3} = 8 \\[1em] \Rightarrow r = \sqrt[3]{8} \\[1em] \Rightarrow r = 2.\end{aligned}$$

Substitute value of r in equation (1), we get:

$$\begin{aligned}\Rightarrow a \times 2^5 = 32 \\[1em] \Rightarrow a \times 32 = 32 \\[1em] \Rightarrow a = \frac{32}{32} \\[1em] \Rightarrow a = 1.\end{aligned}$$

Hence, the first term = 1 and common ratio = 2.

(b) By formula,

$S_n = \frac{a(r^n - 1)}{r - 1}$

Substituting values we get :

$$\begin{aligned}\Rightarrow S_{10} = \frac{1(2^{10} - 1)}{2 - 1} \\[1em] \Rightarrow S_{10} = \frac{1024 - 1}{1} \\[1em] \Rightarrow S_{10} = 1023.\end{aligned}$$

Hence, sum of the first 10 terms of the G.P. = 1023.