Question

How many terms of G.P. $3,3^2,3^3,$ ............ are needed to give the sum 120 ?

Answer 1

Here a = 3 and $r=\frac{3^3}{3}=3$

Let n be the needed terms even that their sum is 120.

i.e, $S_n$ = 120

Since $S_n=\frac{a(r^n-1)}{r-1}$, when r > 1

$\therefore 120=\frac{3(3^n-1)}{3-1}$

$\Rightarrow 120=\frac{3}{2}(3^n-1)$

$\Rightarrow 120\times \frac{2}{3}=3^n-1$

$\Rightarrow 3^n=81\Rightarrow 3^n=(3{)}^4\Rightarrow n=4$

Thus, the sum of 4 terms of the given G.P. is 120.