Question

Consider an infinite geometric series with the first term a and common ratio r. If its sum is 4 and the second term is 3/4, then

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

Answer 1

The correct option is D $a=3$, $r=\frac{1}{4}$

if the first term of the series is $a$ and common ratio is $r$ then the sum of infinite geometric series will be

$\frac{a}{1-r}$

and the series will be $a,ar,a{r}^2,a{r}^3,.........$

given sum$=4$ and second term $=\frac{3}{4}$

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

$a=4-4r$ .......................(1)

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

putting value of $a$ from equation (1)

$(4-4r)r=\frac{3}{4}$

$4(4r-4r^2)=3$

$16r^2-16r+3=0$

$16r^2-12r-4r+3=0$

$4r(4r-3)-1(4r-3)=0$

$(4r-1)(4r-3)=0$

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

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

$a=1$

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

$a=3$