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The sum of n terms of two APs is in the ratio$\; 5n + 1: 4n+10$. Find the ratio of their $\;5^{th}$ terms.

$(a)\;\frac{47}{39}\qquad(b)\;1\qquad(c)\;\frac{6}{30}\qquad(d)\;\frac{5}{4}$

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A)
Answer : (b) 1
Explanation : $\frac{S_{1}}{S_{2}}=\frac{n/2\;[2a_{1}+(n-1)d_{1}]}{n/2\;[2a_{2}+(n-1)d_{2}]}$
$\frac{S_{1}}{S_{2}}=\frac{\;[2a_{1}+\frac{(n-1)}{2}d_{1}]}{\;[2a_{2}+\frac{(n-1)}{2}d_{2}]}=\frac{5n+1}{4n+10}$
For the ratio of the $\;5^{th}\;$ term , $\;\frac{a_{1}+4d_{1}}{a_{2}+4d_{2}}$
Let $\;\frac{n-1}{2}=4\quad\;then\;n=8+1=9$
Replacing the value of n in the ratio ,
$\frac{5*9+1}{4*9+10}=\frac{47}{47}=1.$
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