Question

Consider an infinite geometric series with first term $a$ and common ratio $r$. If its sum is $4$ and the second term is $\frac{3}{4}$, 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 $a=3,r=\frac{1}{4}$

Sum is $4$ and second term is $\frac{3}{4}$.
Let the first term be $a$ and common ratio be $r$,
$\frac{a}{1-r}=4ar=\frac{3}{4}\Rightarrow r=\frac{3}{4a}$
Therefore,
$\frac{a}{1-\frac{3}{4a}}=4\Rightarrow \frac{4a^2}{4a-3}=4\Rightarrow a^2-4a+3=0\Rightarrow (a-1)(a-3)=0\Rightarrow a=1,3$

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