A) \[2a-d=0\]
B) \[a-d=0\]
C) \[a-2d=0\]
D) None of these
Correct Answer: A
Solution :
\[\frac{{{S}_{kn}}}{{{S}_{n}}}=\frac{(kn/2)\{2a+(kn-1)d\}}{(n/2)\{2a+(n-1)d\}}=k\left\{ \frac{(2a-d)+knd}{(2a-d)+nd} \right\}\] \[i.e.\] if\[2a-d=0\], then this becomes \[\frac{{{k}^{2}}nd}{nd}={{k}^{2}}\] which is obviously independent of \[n\].
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