Let S_n and s_n denote the sum of the first n | vedclass

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(C) Given the ratio of the sum of $n$ terms of two $A.P.s$ is $\frac{s_n}{S_n} = \frac{3n - 13}{7n + 13}$.Let the sum of $n$ terms be $s_n = k(3n^2 -

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$\frac{3n - 13}{28n + 26}$

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(C) Given the ratio of the sum of $n$ terms of two $A.P.s$ is $\frac{s_n}{S_n} = \frac{3n - 13}{7n + 13}$.Let the sum of $n$ terms be $s_n = k(3n^2 - 13n)$ and $S_n = k(7n^2 + 13n)$ for some constant $k$.To find the sum of $2n$ terms for the second $A.P.$,we substitute $2n$ for $n$ in the expression for $S_n$:$S_{2n} = k(7(2n)^2 + 13(2n)) = k(7(4n^2) + 26n) = k(28n^2 + 26n)$.Therefore,the ratio $\frac{s_n}{S_{2n}} = \frac{k(3n^2 - 13n)}{k(28n^2 + 26n)} = \frac{3n^2 - 13n}{28n^2 + 26n} = \frac{n(3n - 13)}{n(28n + 26)} = \frac{3n - 13}{28n + 26}$.

Let $S_n$ and $s_n$ denote the sum of the first $n$ terms of two different arithmetic progressions $(A.P.)$ for which $\frac{s_n}{S_n} = \frac{3n - 13}{7n + 13}$. Find the ratio $\frac{s_n}{S_{2n}}$.

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