Solve the following

Question:

If, $S_{1}$ is the sum of an arithmetic progression of ' $n$ ' odd number of terms and $S_{2}$ 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) $\frac{n+1}{n}$

Solution:

(a) $\frac{2 n}{n+1}$

Let n be an odd number.

Given:

$S_{1}=$ Sum of odd number of terms

$=\frac{n}{2}\{2 a+(n-1) d\} \quad \ldots .(1)$

Since $n$ is odd, the number of odd places $=\frac{n+1}{2}$

$S_{2}=$ Sum of the terms of a series in odd places

$=\frac{\left(\frac{n+1}{2}\right)}{2}\left\{2 a+\left(\frac{n+1}{2}-1\right) 2 d\right\}$

$=\frac{n+1}{4}\{2 a+(n-1) d\} \quad \ldots(2)$

From equations $(1)$ and $(2)$, we have:

$\frac{S_{1}}{S_{2}}=\frac{\frac{n}{2}\{2 a+(n-1) d\}}{\frac{n+1}{4}\{2 a+(n-1) d\}}$

$=\frac{2 n}{n+1}$

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