Question

The sum of the first and the last term of an increasing geometric progression is $66$, the product of the second and the last but one term is $128$ and the sum of all its terms is $126$. How many terms are there in the progression?

Answer 1

Let the G.P. be $a,ar,a{r}^2,.......,a{r}^n$

Given that:-

$a+a{r}^n=66.....\left(1\right)$

$a.a{r}^n=128$

$\Rightarrow r^n=\frac{128}{a^2}.....\left(2\right)$

Substituting the value of $r^n$ in equation $\left(1\right)$, we have

$a+a\left(\frac{128}{a^2}\right)=66$

$\Rightarrow a^2+128=66a$

$\Rightarrow a^2-66a+128=0$

$\Rightarrow a^2-64a-2a+128=0$

$\Rightarrow (a-64)(a-2)=0$

$\Rightarrow a=64,2$

As the G.P. is increasing, i.e., $r>1$

$\therefore a=2$

Substituting the value of $a$ in ${eq}^n\left(2\right)$, we get

$r^n=\frac{128}{2^2}=32.....\left(3\right)$

Also given that sum of $n$ terms of G.P. is 126.

$\therefore \frac{2(r^{n+1}-1)}{r-1}=126$

$\Rightarrow \frac{32r-1}{r-1}=\frac{126}{2}[\because r^{n+1}=r^n.r]$

$\Rightarrow 32r-1=63r-63$

$\Rightarrow 31r=62$

$\Rightarrow r=2$

Substituting the value of $r$ in ${eq}^n\left(3\right)$, we get

$2^n=32$

$\Rightarrow 2^n=2^5$

$\Rightarrow n=5$

Hence the correct answer is $5$.