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(A) The sum of the first $k$ terms of an arithmetic progression is given by $S_k = \frac{k}{2} \{2a + (k - 1)d\}$.Therefore,the ratio $\frac{S_{kn}}{S

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$2a - d = 0$

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(A) The sum of the first $k$ terms of an arithmetic progression is given by $S_k = \frac{k}{2} \{2a + (k - 1)d\}$.Therefore,the ratio $\frac{S_{kn}}{S_n}$ is given by:$\frac{S_{kn}}{S_n} = \frac{\frac{kn}{2} \{2a + (kn - 1)d\}}{\frac{n}{2} \{2a + (n - 1)d\}}$Simplifying the expression:$= k \left\{ \frac{2a + knd - d}{2a + nd - d} \right\} = k \left\{ \frac{(2a - d) + knd}{(2a - d) + nd} \right\}$For this expression to be independent of $n$,the term $(2a - d)$ must be equal to $0$.If $2a - d = 0$,then the expression becomes:$= k \left\{ \frac{knd}{nd} \right\} = k^2$Since $k^2$ is independent of $n$,the condition is $2a - d = 0$.

If $S_k$ denotes the sum of the first $k$ terms of an arithmetic progression whose first term and common difference are $a$ and $d$ respectively,then $S_{kn}/S_n$ is independent of $n$ if

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