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

(a) first term and the common ratio.

(b) sum of its first 10 terms.

(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:

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

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

$\Rightarrow a \times 2^5 = 32 \\[1em] \Rightarrow a \times 32 = 32 \\[1em] \Rightarrow a = \dfrac{32}{32} \\[1em] \Rightarrow a = 1.$

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

(b) By formula,

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

Substituting values we get :

$\Rightarrow S${10} = \dfrac{1(2^{10} - 1)}{2 - 1} \\[1em] \Rightarrow S{10} = \dfrac{1024 - 1}{1} \\[1em] \Rightarrow S_{10} = 1023.

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