Let $a_1, a_2, a_3, \ldots ., a_{100}$ be an arithmetic progression with $a_1=3$ and $S_p=\sum_{i=1}^p a_i$,…
Let $a_1, a_2, a_3, \ldots ., a_{100}$ be an arithmetic progression with $a_1=3$ and $S_p=\sum_{i=1}^p a_i$, $1 \leq p \leq 100$. For any integer $n$ with $1 \leq n \leq 20$, let $m=5 n$. If $\frac{S_m}{S_n}$ does not depend on $n$, then $a_2$ is
Solution
Given, $a_1=3, m=5 n$ and $a_1, a_2, \ldots$ are in AP. $\therefore \frac{S_m}{S_n}=\frac{S_{5 n}}{S_n}$ is independent of $n$. Now, $\quad \frac{\frac{5 n}{2}[2 \times 3+(5 n-1) d]}{\frac{n}{2}[2 \times 3+(n-1) d]}$ $\Rightarrow \quad \frac{f\{(6-d)+5 n\}}{(6-d)+n}$ independent of $n$, if $\begin{aligned} \quad 6-d= & 0 \Rightarrow d=6 \\ \therefore \quad a_2= & a_1+d=3+6=9 \\ & \text { Or }\end{aligned}$
If $d=0$ $\frac{S_m}{S_n}$ is independent of $n$.
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\therefore \quad a_2=3
$