Question

If 1701 is the n^th term of the Geometric Progression (G.P.) 7, 21, 63 ……., find:

1. the value of ‘n’
2. hence find the sum of the ‘n’ terms of the G.P

Answer 1

(a) Given, G.P. 7, 21, 63, ........,

n^th term (Tn) = 1701

First term (a) = 7

Common ratio (r) = `21/7 = 3`

Tⁿ = 1701

ar^{n −1} = 1701

`7 × (3)^(n −1) = 1701`

`(3)^(n − 1) = 1701/7`

= 243

`(3)^(n − 1) = (3)^5`

n − 1 = 5

n = 5 + 1 = 6

(b) Sum of first 6 terms

`S_n = (a(r^n − 1))/(r − 1)`

`S_6 = (7(3^6 − 1))/(3 − 1)`

`S_6 = (7(729 − 1))/(3 − 1)`

`S_6 = (7 xx 728)/2`

S₆ = 2548