Question

What is the sum of the first five terms of a geometric series with`" "a_1 = 6 and r = 1/3`?

Answer 1

`242/27`

Explanation:

We know sum of n terms of G.P. IS, `s_n=[a(1-r^n)]/[1-r]`

Put n=5,a=6 and r = 1/3 we get

`rArr S_5=[6{1-(1/3)^5}]/[(1-1/3)]`

`rArr S_5= [6*242/243]/(2/3)`

`rArr S_5= 6*242/243*3/2 = 242/27`