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If the number of terms of an A.P. is $(2 n+1)$, then what is the ratio of the sum of the odd terms to the sum of even terms?
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The correct answer is:
$\frac{n+1}{n}$
Let the $\mathrm{AP}$ is $\mathrm{a}, \mathrm{a}+\mathrm{d}, \mathrm{a}+2 \mathrm{~d}, \ldots \ldots, \mathrm{a}+(2 \mathrm{n}-1) \mathrm{d}, \mathrm{a}+2 \mathrm{nd}$
Series of even terms. $\mathrm{a}+\mathrm{d}, \mathrm{a}+3 \mathrm{~d}, \ldots \ldots \ldots, \mathrm{a}+(2 \mathrm{n}-1) \mathrm{d}$, has $\mathrm{n}$ terms
$\begin{aligned} \text { Sum of even number } &=\frac{n}{2}[(a+d)+\{a+(2 n-1) d\}] \\ &=\frac{n}{2}[2 a+2 n d]=n[a+n d] \end{aligned}$
Series of odd terms $\mathrm{a}, \mathrm{a}+2 \mathrm{~d}, \mathrm{a}+4 \mathrm{~d}, \ldots \ldots \ldots \ldots, \mathrm{a}+2 \mathrm{nd}$, has $(\mathrm{n}+1)$ terms.
$\begin{aligned} \text { Sum of odd numbers } &=\frac{n+1}{2}[a+(a+2 n d)] \\ &=\frac{n+1}{2}(2 a+2 n d) \\ &=(n+1)(a+n d) \end{aligned}$
So, the required ratio $=\frac{n+1}{n}$
Series of even terms. $\mathrm{a}+\mathrm{d}, \mathrm{a}+3 \mathrm{~d}, \ldots \ldots \ldots, \mathrm{a}+(2 \mathrm{n}-1) \mathrm{d}$, has $\mathrm{n}$ terms
$\begin{aligned} \text { Sum of even number } &=\frac{n}{2}[(a+d)+\{a+(2 n-1) d\}] \\ &=\frac{n}{2}[2 a+2 n d]=n[a+n d] \end{aligned}$
Series of odd terms $\mathrm{a}, \mathrm{a}+2 \mathrm{~d}, \mathrm{a}+4 \mathrm{~d}, \ldots \ldots \ldots \ldots, \mathrm{a}+2 \mathrm{nd}$, has $(\mathrm{n}+1)$ terms.
$\begin{aligned} \text { Sum of odd numbers } &=\frac{n+1}{2}[a+(a+2 n d)] \\ &=\frac{n+1}{2}(2 a+2 n d) \\ &=(n+1)(a+n d) \end{aligned}$
So, the required ratio $=\frac{n+1}{n}$
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