Question

In a geometric progression, the sum of first n terms is 65535. If the last term is 49152 and the common ratio is 4, then find the value of n.

Answer 1

Let the common ratio and the first term of the GP be r and a respectively. Given, $r=4$ and $a{r}^{n-1}=49152⟹a{r}^n=49152r⟹a{r}^n=49152\times 4$ ...(1)
Also, $\frac{a(r^n-1)}{r-1}=65535⟹\frac{a(r^n-1)}{3}=65535$
$⟹a{r}^n-a=3\times 65535⟹49152\times 4-3\times 65535=a⟹a=3$.
Substituting $a=3$ in $a{r}^{n-1}=49152$, we get $n=8$.