Question

In a $G.P.$ the sum of the first and last terms is $66$ , the product of the second and the last but one is $128$ , and the sum of the terms is $126$.
If the decreasing $G.P.$ is considered, then the sum of infinite terms is

• A. 64
• B. 128
• C. 256
• D. 729

Answer 1

Answer: (B)

Let a be the first term and $r$ the common ratio of the given $G.P.$
Further, let there be $n$ terms in the given G.P. Then
$a_1+a_n=66$
$\Rightarrow a+a{r}^{n-1}=66...(i)$
$a_2\times a_{n-1}=128$
$\Rightarrow ar\times a{r}^{n-2}=128$
or $a^2{v}^{n-1}=128$
or $a\times \left(a{r}^{n-1}\right)=128$
or $a{r}^{n-1}=\frac{128}{a}$
Putting this value of $a{r}^{n-1}$ in (i), we get
$a+\frac{128}{a}=66$
or $a^2-66a+128=0$
or $(a-2)(a-64)=0$
or $a=2,64$
Putting $a=2$ in (i), we get
$2+2\times 1^{n-1}=66$
or $r^{n-1}=32$
Putting $a=64$ in (i), we get
$64+641^{n-1}=66$
or $r^{n-1}=\frac{1}{32}$
For an increasing G.P. $r>1$.
Now, $S_n=126$
$\Rightarrow 2\left(\frac{r^n-1}{r-1}\right)=126$
or $\frac{r^n-1}{r-1}=63$
or $\frac{r^{n-1}\times r-1}{r-1}=63$
or $\frac{32r-1}{r-1}=63$
or $r=2$
$\therefore 1^{n-1}=32$
$\Rightarrow 2^{n-1}=32=2^5$
$\Rightarrow n-1=5$
$\Rightarrow n=6$
For decreasing G.P., $a=64$ and $r=1/2$.
Hence, the sum of infinite terms is $64/\{1-(1/2)\}=128.$
For $a=2,r=2$, terms are $2,4,8,16,32,64$
For $a=64,r=1/2$ terms are $64,32,16,8,4,2$.
Hence, difference is $62$ .