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 $\left(a,r\right)$ is:

A

$\left(\frac{4}{7},\frac{3}{7}\right)$

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B

$\left(2,\frac{3}{8}\right)$

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C

$\left(\frac{3}{2},\frac{1}{2}\right)$

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D

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

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Solution

## 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}\\ ⇒r& =& \frac{3}{4a}\end{array}$Substituting the value of $r$ in $\left(1\right)$,$\begin{array}{rcl}\frac{a}{1-\frac{3}{4a}}& =& 4\\ ⇒\frac{a}{\frac{4a-3}{4a}}& =& 4\\ ⇒a×\frac{4a}{4a-3}& =& 4\\ ⇒4{a}^{2}& =& 16a-12\\ ⇒{a}^{2}& =& 4a-3\\ ⇒{a}^{2}-4a+3& =& 0\\ ⇒{a}^{2}-a-3a+3& =& 0\\ & ⇒& a\left(a-1\right)-3\left(a-3\right)=0\\ ⇒\left(a-1\right)\left(a-3\right)& =& 0\end{array}$When $a=3$, $r=\frac{1}{4}$ and when $a=1$, $r=\frac{3}{4}$ Hence, option D is correct.

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