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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:
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Verified Answer
The correct answer is:
$a=3, r=\frac{1}{4}$
Since, sum $=4$ and second term $=\frac{3}{4}$ $\Rightarrow \frac{a}{1-r}=4$, and $a r=\frac{3}{4}$ $\Rightarrow \frac{a}{1-\frac{3}{4 a}}=4$ $\Rightarrow(a-1)(a-3)=0$ $\Rightarrow a=1$ or $a=3$
$\begin{array}{l}
\Rightarrow \frac{a}{1-\frac{3}{4 a}}=4 \\
\Rightarrow(a-1)(a-3)=0 \\
\Rightarrow a=1 \text { or } a=3
\end{array}$
$\begin{array}{l}
\Rightarrow \frac{a}{1-\frac{3}{4 a}}=4 \\
\Rightarrow(a-1)(a-3)=0 \\
\Rightarrow a=1 \text { or } a=3
\end{array}$
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