Question

Let $a_{1,}{a}_{2,...,}{a}_{100}$be in Arithmetic progression with $a_1=3$ and $S_p=\sum _{i=1}^p{a}_i,1\le p\le 100$. Find any integer $n$with $1\le n\le 20$, let $m=5n$.If $\frac{S_m}{S_n}$ does not depend on $n$, then $a_2$ is

• A. $9$
• B. $2\text{or}4$
• C. $4\text{or}16$
• D. None of these

Answer 1

The correct option is A

$9$

Explanation for the correct answer:

Step1: Determining the value of $d$:

An arithmetic progression (AP) is a progression in which the difference between two consecutive terms is constant.

Sum of $n$ terms of AP

$S_n=\frac{n}{2}\left[2a_1+\left(n-1\right)d\right]$ where $a_1=\text{first term},d=\text{common difference}$.

Given $S_p=\sum _{i=1}^p{a}_i,1\le p\le 100$

$$\begin{gathered} \frac{S_m}{S_n}=\frac{\frac{m}{2}\left[2\times 3+\left(m-1\right)d\right]}{\frac{n}{2}\left[2\times 3+\left(n-1\right)d\right]}\mathbf{[}\mathbf{\because }{\mathit{a}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{}\mathit{G}\mathit{i}\mathit{v}\mathit{e}\mathit{n}\mathbf{]} \\ =\frac{\frac{5n}{2}\left[2\times 3+\left(5n-1\right)d\right]}{\frac{n}{2}\left[2\times 3+\left(n-1\right)d\right]}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{m}\mathbf{=}\mathbf{5}\mathbf{n}\mathbf{,}\mathbf{}\mathbf{G}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{]} \\ =\frac{5\left[6+\left(5n-1\right)d\right]}{\left[6+\left(n-1\right)d\right]} \\ =5\frac{\left[6-d+5nd\right]}{\left[6-d+nd\right]} \end{gathered}$$

Since $\frac{S_m}{S_n}$ does not depend on $n$

$$\begin{gathered} \Rightarrow 6-d=0 \\ \Rightarrow d=6 \end{gathered}$$

Step-2: finding the value of $a_2$

$$\begin{gathered} a_n=a_1+\left(n-1\right)d \\ =3+\left(2-1\right)6 \\ {a}_2=9 \end{gathered}$$

Hence,the correct answer is option (A).