Question

Consider an infinite $G.P$. with first term $a$ and common ratio $r$, its sum is $4$ and the second term is $\frac{3}{4}$, then?

• A. $a=\frac{4}{7},r=\frac{3}{7}$
• B. $a=\frac{3}{2},r=\frac{1}{2}$
• C. $a=1,r=\frac{3}{4}$
• D. $a=3,r=\frac{1}{4}$

Answer 1

Answer: (C), (D)

$a=1,r=\frac{3}{4}$
$a=3,r=\frac{1}{4}$

Given:-

$S_{\infty }=4$

$a_2=\frac{3}{4}$

$\Rightarrow ar=\frac{3}{4}$

$\Rightarrow 4ar=3.....\left(1\right)$

As we know that,

$S_{\infty }=\frac{a}{1-r}$

$\therefore \frac{a}{1-r}=4$

$\Rightarrow 4r=4-a.....\left(2\right)$

From equation $\left(1\right)\&\left(2\right)$, we have

$a(4-a)=3$

$\Rightarrow a^2-4a+3=0$

$\Rightarrow (a-3)(a-1)=0$

$\Rightarrow a=1$ or $a=3$

Substituting the value of $a$ in equation $\left(1\right)$, we get

For $a=3$

$\Rightarrow r=\frac{1}{4}$

For $a=1$

$\Rightarrow r=\frac{3}{4}$