Question

In a GP the product of the first four terms is $4$ and the second term is the reciprocal of the fourth term. The sum of the GP up to infinite terms is

• A. $8$
• B. $-8$
• C. $\frac{8}{3}$
• D. $-\frac{8}{3}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $8$
B $-8$
C $\frac{8}{3}$
D $-\frac{8}{3}$

Let the first four terms of the GP be $\frac{a}{r^3},\frac{a}{r},ar,a{r}^3$ with common ratio $r^2$.
$\therefore a^4$=4 and $\frac{a}{r}=\frac{1}{a{r}^3}$
$\Rightarrow a=4^{1/4}$ and $a^2{r}^2=1$
$\therefore a=\pm \sqrt{2},\pm \sqrt{2}i$ and $ar=\pm 1$
Case(i) Taking $a=\sqrt{2}$ and $r=-\frac{1}{\sqrt{2}}$
The first four terms of GP are -4,-2,-1,-$\frac{1}{2}$ with common ratio=$\frac{1}{2}$
$\therefore S_{\infty }=\frac{-4}{1-\frac{1}{2}}$

$=-8$

$\therefore$ Option B is correct.
Case(ii) Taking $a=\sqrt{2}$ and $r=\frac{1}{\sqrt{2}}$
The first four terms of GP are 4,2,1,$\frac{1}{2}$ with common ratio=$\frac{1}{2}$
$\therefore S_{\infty }=\frac{4}{1-\frac{1}{2}}$

$=8$

$\therefore$ Option A is correct.
Case(iii) Taking $a=\sqrt{2}i$ and $r=\frac{1}{\sqrt{2}i}$
The first four terms of GP are 4,-2,1,$\frac{-1}{2}$ with common ratio=$\frac{-1}{2}$
$\therefore S_{\infty }=\frac{4}{1-(-\frac{1}{2})}$

= $\frac{8}{3}$

$\therefore$ Option C is correct.
Case(iv) Taking $a=\sqrt{2}i$ and $r=-\frac{1}{\sqrt{2}i}$
The first four terms of GP are -4,2,-1,$\frac{1}{2}$ with common ratio=$\frac{-1}{2}$
$\therefore S_{\infty }=\frac{-4}{1-(-\frac{1}{2})}$

= -$\frac{8}{3}$

$\therefore$ Option D is correct.