If S1 is the sum of an arithmetic progression of ‘n’ odd number of terms and S2 is the sum of the terms of the series in odd places, then S_1S_2

Answer: Option A

Odd numbers are 1, 3, 5, 7, 9, 11, 13, ...... n
∴ S₁ = Sum of odd numbers = n²
S₂ = Sum of number at odd places
3, 7, 11, 15, ......
a = 3, d = 7 - 3 = 4 and number of term = $$\frac{n}{2}$$
$$\begin{array}{rl}& S_2=\frac{n}{2\times 2}\left[2\times 3+\left(\frac{n}{2}-1\right)\times 4\right]\\ & =\frac{n}{4}\left[6+2n-4\right]\\ & =\frac{n}{4}\left[2n+2\right]\\ & =\frac{n\left(n+1\right)}{2}\\ & \therefore \frac{s_1}{s_2}=\frac{n^2\times 2}{n\left(n+1\right)}\\ & =\frac{2n}{n+1}\end{array}$$