Question

Consider an AP with first term a and the common difference $d$. Let $S_k$ denote the sum of its first $k$ terms, if $\frac{S_{kn}}{S_n}$ is independent of $n$ then:

• A. $2a-d=0$
• B. $a-d=0$
• C. $a-2d=0$
• D. None of these

Answer 1

The correct option is A

$2a-d=0$

Explanation for the correct option

Given, first term $=a$, common difference$=d$
We know that, sum of n terms in A.P is ${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2\mathrm{a}+(\mathrm{n}-1\left)\mathrm{d}\right]$ we have,
$$\begin{gathered} \Rightarrow \frac{S_{kn}}{S_n}=\frac{\frac{kn}{2}(2a+(kn-1\left)d\right)}{\frac{n}{2}(2a+(n+1\left)d\right)} \\ \Rightarrow \frac{S_{kn}}{S_n}=\frac{\mathrm{k}(2\mathrm{a}+(kn-1\left)\mathrm{d}\right)}{(2\mathrm{a}+(n+1\left)\mathrm{d}\right)} \\ \Rightarrow \frac{S_{kn}}{S_n}=\frac{k(2a-d+knd)}{2a+nd-d} \end{gathered}$$
For independent of $n$ only when $2a=d$ or $2a-d=0$

Hence, option A is correct.