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

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Solution

Let a and d be the first term and common difference respectively of the given AP.

$a_{k} = a + (k - 1) d$

Now,
$S_{1} = S u m o f o d d t e r m s$

$S_{1} = a_{1} + a_{3} + a_{5} + . . . . + a_{2 n + 1}$

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

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

$S_{1} = (n + 1) (a + n d)$

And,
$S_{2} = S u m o f e v e n t e r m s$

$S_{2} = a_{2} + a_{4} + a_{6} + . . . . . + a_{2 n}$

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

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

$S_{2} = n (a + n d)$

Therefore,
$S_{1} : S_{2} = (n + 1) (a + n d) : n (a + n d) = (n + 1) : n$

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