Question

In an infinite geometric series, the first term is $a$ and the common ratio is $r$. If the sum of the series is $4$ and the second term is $\frac{3}{4}$, then $(a,r)$ is:

• A. $\left(\frac{4}{7},\frac{3}{7}\right)$
• B. $\left(2,\frac{3}{8}\right)$
• C. $\left(\frac{3}{2},\frac{1}{2}\right)$
• D. $\left(3,\frac{1}{4}\right)$

Answer 1

The correct option is D

$\left(3,\frac{1}{4}\right)$

Explanation for the correct option:

Finding the value of $\left(a,r\right)$:

Sum of infinite terms of GP is $\frac{a}{1-r}$.

$\frac{a}{1-r}=4...\left(1\right)$

Second term of GP $=ar$

$\begin{array}{rcl}ar& =& \frac{3}{4}\\ \Rightarrow r& =& \frac{3}{4a}\end{array}$

Substituting the value of $r$ in $\left(1\right)$,

$\begin{array}{rcl}\frac{a}{1-{\displaystyle \frac{3}{4a}}}& =& 4\\ \Rightarrow \frac{a}{{\displaystyle \frac{4a-3}{4a}}}& =& 4\\ \Rightarrow a\times \frac{4a}{4a-3}& =& 4\\ \Rightarrow 4a^2& =& 16a-12\\ \Rightarrow a^2& =& 4a-3\\ \Rightarrow a^2-4a+3& =& 0\\ \Rightarrow a^2-a-3a+3& =& 0\\ & \Rightarrow & a\left(a-1\right)-3\left(a-3\right)=0\\ \Rightarrow \left(a-1\right)\left(a-3\right)& =& 0\end{array}$

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

Hence, option D is correct.