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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 $\frac{S_{1}}{S_{2}}$.

A. $\frac{2 n}{n+1}$ B. $\frac{n}{n+1}$ C. $\frac{n+1}{2 n}$ D. $\frac{n-1}{n}$ Answer: Option A
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Solution(By Apex Team)

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 = $\Large\frac{n}{2}$ $\begin{aligned}S_2&=\frac{n}{2\times2}\left[2\times3+\left(\frac{n}{2}-1\right)\times4\right]\\ &=\frac{n}{4}[6+2n-4]\\ &=\frac{n}{4}[2n+2]\\ &=\frac{n(n+1)}{2}\\ \therefore&\frac{s_1}{s_2}=\frac{n^2\times2}{n(n+1)}\\ &=\frac{2n}{n+1}\end{aligned}$

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