Question

The first and last terms of a Geometrical Progression (G.P.) are 3 and 96, respectively. If the common ratio is 2, find:

1. ‘n’ the number of terms of the G.P.
2. Sum of the n terms.

Answer 1

Given a = 3,  l = 96

(i) r = 2 

l = ar^{n -1}

96 = 3 (2) ^n-1 

⇒ 32 = (2)^{n -1}

⇒ 2⁵ = 2 ^n-1

∴ n = 6

(ii) Sum of the n terms = (S_n) 

` = (a(r^n - 1))/(r - 1)`

`=(3(2^6 - 1))/(2 - 1)` 

= 3 × 63

= 189