If there are $2 n + 1$ terms in A.P., then prove that the ratio of the sum of odd terms and even terms is $(n + 1) : n$

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Solution

$a_{1} + a_{2} + a_{3} + . . . . . a_{2 n + 1}$

We have $n + 1$ odd term and n even terms

For odd terms $n = n + 1$

$d = 2 d$

$a = a_{1}$

$S_{n_{o d d}} = \frac{n + 1}{2} [2 a_{1} + (n + 1 - 1) 2 d]$

$= \frac{n + 1}{2} [2 a_{1} + 2 n d]$

$= (n + 1) [a_{1} + n d]$

For even terms $a = a_{1} + d$

$n = n$

$d = 2 d$

$S_{n_{e v e n}} = \frac{n}{2} [2 (a_{1} + d) + (n - 1) 2 d]$

$= \frac{n}{2} [2 a_{1} + 2 n d]$

$= n [a_{1} + n d]$

$\frac{S_{n_{o d d}}}{S_{n_{e v e n}}} = \frac{n + 1}{n}$

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