If {S_k} denotes the sum of the first k terms | vedclass

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

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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 ${S_k} = \frac{k}{2} \{2a + (k - 1)d\}$.We are given the ratio $\frac{S_{kn}}{S_n} = \frac{\frac{kn}{2} \{2a + (kn - 1)d\}}{\frac{n}{2} \{2a + (n - 1)d\}}$.Simplifying the expression,we get:$\frac{S_{kn}}{S_n} = k \left\{ \frac{2a + (kn - 1)d}{2a + (n - 1)d} \right\} = k \left\{ \frac{(2a - d) + knd}{(2a - d) + nd} \right\}$.For this expression to be independent of $n$,the term containing $n$ in the numerator must be proportional to the term containing $n$ in the denominator,or the constant terms must vanish such that $n$ cancels out.If $2a - d = 0$,then the expression becomes:$\frac{S_{kn}}{S_n} = k \left\{ \frac{knd}{nd} \right\} = k^2$.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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