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Given that $\frac{S_{1}}{S_{2}}=\frac{3n+8}{7n+5}$, $\frac{S_{1}}{S_{2}}$ is the ratio of sum of two APs to n terms, find the ratio of the 11<sup>th</sup> terms of the two APs.

$\begin{array}{1 1}\frac{72}{151} \\ \frac{71}{152} \\ \frac{152}{72} \\ \frac{151}{71}\end{array}$

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Answer : (b) 71/152
Explanation : $\frac{S_{1}}{S_{2}}=\frac{3n+8}{7n+5}$
$\frac{S_{1}}{S_{2}}=\frac{(2a_{1}+(n-1)d_{1})n/2}{(2a_{2}+(n-1)d_{2})n/2}=\frac{a_{1}+(n-1)/2\;d_{1}}{a_{2}+(n-1)/2\;d_{2}}$
$We \;require\; the\; ratio \;of\; 11th\; terms \;\frac{A_{1}}{A_{2}}=\frac{a_{1}+10d_{1}}{a_{2}+10d_{2}}$
$Replacing\;\frac{n-1}{2}=10\;in\;\frac{S_{1}}{S_{2}}$
$We\; get\; n=21$
$\frac{A_{1}}{A_{2}}=\frac{3*21+8}{7*21+5}=\frac{71}{152}.$
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