Question

Consider an infinite geometric series with first term ${}^{\prime }{a}^{\prime }$ and common ratio ${}^{\prime }{r}^{\prime }$. If the sum is $4$ and the second term is $\frac{3}{4}$, then

• A. $a=\frac{4}{7},r=\frac{3}{7}$
• B. $a=3,r=\frac{1}{4}$
• C. $a=2,r=\frac{3}{8}$
• D. $a=\frac{3}{2},r=\frac{1}{2}$

Answer 1

Answer: (B)

$\because$ Sum of infinite geometric series,
$S_{\infty }=\frac{a}{1-r}$
$\Rightarrow 4=\frac{a}{1-r}\Rightarrow a=4-4r$...(i)
Also, $T_2=ar\Rightarrow ar=\frac{3}{4}$...(ii)
$\therefore$ From Eqs. (i) and (ii), we get
$a=4-4\left(\frac{3}{4a}\right)$
$\Rightarrow a=4-\frac{3}{a}$
$\Rightarrow a^2-4a+3=0$
$\Rightarrow a^2-3a-a+3=0$
$\Rightarrow a(a-3)-1(a-3)=0$
$\Rightarrow a=1,3$
On putting $a=3$ in Eq. (ii), we get
$3r=\frac{3}{4}$
$\Rightarrow r=\frac{1}{4}$