Let $a_1, a_2, a_3, \ldots, a_{100}$ be an arithmetic progression with $a_1 = 3$ and $S_p = \sum_{i=1}^{p} a_i$, $1 \le p \le 100$. For 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

Let a₁, a₂, a₃, …, a₁₀₀ be an arithmetic progression with a₁ = 3 and Sₚ = ∑ᵢ₌₁ᵖ aᵢ, 1 ≤ p ≤ 100. For any integer n with 1 ≤ n ≤ 20, let m = 5n. If Sₘ/Sₙ does not depend on n, then a₂ is $\text{Given: } a_1, a_2, \ldots, a_{100} \text{ is an A.P., } a_1 = 3$ $a_n = a_1 + (n-1)d, \qquad S_p = \sum_{i=1}^{p} a_i, \quad 1 \leq p \leq 100$ $\text{To find: } a_2, \quad \text{given } m = 5n, \ 1 \leq n \leq 20$ $\dfrac{S_m}{S_n} = \dfrac{S_{5n}}{S_n} \text{ must be the same for every } n$ $1.\ \text{write } S_p \text{ for this A.P.}$ $2.\ \text{form } \dfrac{S_{5n}}{S_n}$ $3.\ \text{force it independent of } n; \text{ find } d$ $S_p = \dfrac{p}{2}\left[2a_1 + (p-1)d\right] = \dfrac{p}{2}\left[6 + (p-1)d\right]$ $S_{5n} = \dfrac{5n}{2}\left[6 + (5n-1)d\right]$ $S_n = \dfrac{n}{2}\left[6 + (n-1)d\right]$ $\dfrac{S_{5n}}{S_n} = \dfrac{\;\frac{5n}{2}\left[6+(5n-1)d\right]\;}{\;\frac{n}{2}\left[6+(n-1)d\right]\;}$ $= \dfrac{5\left[6+(5n-1)d\right]}{6+(n-1)d}$ $= \dfrac{5\left[(6-d) + 5nd\right]}{(6-d) + nd}, \qquad c := 6-d$ $\Rightarrow \dfrac{S_{5n}}{S_n} = \dfrac{5(c+5nd)}{c+nd}$ $\dfrac{S_{5n}}{S_n} \text{ independent of } n \ \text{ for } n = 1,2,\ldots,20$ $\left.\dfrac{S_{5n}}{S_n}\right|_{n=1} = \left.\dfrac{S_{5n}}{S_n}\right|_{n=2}$ $n = 1: \quad \dfrac{S_5}{S_1} = \dfrac{5(6+4d)}{6}$ $n = 2: \quad \dfrac{S_{10}}{S_2} = \dfrac{5(6+9d)}{6+d}$ $\dfrac{5(6+4d)}{6} = \dfrac{5(6+9d)}{6+d} \ \Rightarrow \ \dfrac{6+4d}{6} = \dfrac{6+9d}{6+d}$ $(6+4d)(6+d) = 6(6+9d)$ $36 + 30d + 4d^2 = 36 + 54d$ $4d^2 - 24d = 0$ $4d(d-6) = 0 \ \Rightarrow \ d = 0 \ \text{ or } \ d = 6$ $d = 0 \text{ is the trivial constant A.P.} \ \Rightarrow \ d = 6$ $a_2 = a_1 + d = 3 + 6 = 9$ $\text{Check } (d=6): \ S_n = \dfrac{n}{2}\left[6+(n-1)6\right] = 3n^2$ $\dfrac{S_{5n}}{S_n} = \dfrac{3(5n)^2}{3n^2} = 25 \ \text{ (independent of } n)$