Question

In an increasing $G.P$., the sum of the first and last term is $66$, the product of the second and the last terms is $128$ and the sum of the terms is $126$. Then number of terms in the series is

• A. 10
• B. 6
• C. 8
• D. None of these

Answer 1

Answer: (B)

Let $a$ be the first term, $r$ the common ratio and $n$ be the numbers of terms of the given $G.P$. Then,
$a_{1}+a_{n}=66$
$\Rightarrow a+ a r^{n-1}=66$ ......(1)
$\Rightarrow a_{2} \cdot a_{n-1}=128$
$\Rightarrow a r \cdot a r^{n-2}=128$
$\Rightarrow a r^{2}=128$
$a \cdot\left(a r^{n-1}\right)=128=a r^{n-1}=\frac{128}{a}$
Putting this value of $a r^{n-1}$ in (1), we get $a+\frac{128}{a}=66$
$\Rightarrow a^{2}-66 a+128=0$
$\Rightarrow (a-2)(a-64)=0$
$\Rightarrow a=2,64$
Putting $a=2$ in (1), we get $2+2 \cdot r^{n-1}=66$
$\Rightarrow r^{n-1}=32$
Putting $a=64$ in (1), we get $64+64 r^{n-1}=66$
$\Rightarrow r^{n-1}=\frac{1}{32}$
We reject the second value as the $G.P$. is an increasing $G.P$.
and therefore, $r>1$.
Given, $S_{n}=126$
$\Rightarrow 2\left(\frac{r^{n}-1}{r-1}\right)=126$
$\Rightarrow \frac{r^{n-1} \cdot r-1}{r-1}=63$
$\Rightarrow \frac{32 r-1}{r-1}=63$
$\Rightarrow r=2$
$\therefore r^{n-1}=32$
$\Rightarrow 2^{n-1}=32=2^{5}$
$\Rightarrow n-1=5$
$\Rightarrow n=6$