Question

The sum of the first $n$ terms of the geometric progression, whose first term is $4$ and the common ratio is $3$, is $4372$. Find $n$.

Answer 1

Given that,

$S_n=4372,a=4,r=3$

Sum of $n$ terms of $GP$

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

$S_n(r-1)=a\times 3^n-a$

$4372(3-1)=4\times 3^n-4$

$3^n=\frac{4+8744}{4}$

$3^n=2187$

$3^n=3^7\Rightarrow n=7$

Hence, there are $n=7$ terms in the $GP$.