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प्रश्न
If, S1 is the sum of an arithmetic progression of 'n' odd number of terms and S2 the sum of the terms of the series in odd places, then \[\frac{S_1}{S_2}\] =
पर्याय
\[\frac{2n}{n + 1}\]
\[\frac{n}{n + 1}\]
\[\frac{n + 1}{2n}\]
\[\frac{n + 1}{n}\]
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उत्तर
\[\frac{2n}{n + 1}\]
Let n be an odd number.
Given:
\[S_1 = \text { Sum of odd number of terms }\]
\[ = \frac{n}{2}\left\{ 2a + \left( n - 1 \right)d \right\} . . . . . \left( 1 \right)\]
\[\text { Since n is odd, the number of odd places } = \frac{n + 1}{2}\]
\[ S_2 = \text { Sum of the terms of a series in odd places }\]
\[ = \frac{\left( \frac{n + 1}{2} \right)}{2}\left\{ 2a + \left( \frac{n + 1}{2} - 1 \right)2d \right\}\]
\[ = \frac{n + 1}{4}\left\{ 2a + \left( n - 1 \right)d \right\} . . . . . \left( 2 \right)\]
From equations \[\left( 1 \right) \text { and } \left( 2 \right)\] ,we have:
\[\frac{S_1}{S_2} = \frac{\frac{n}{2}\left\{ 2a + \left( n - 1 \right)d \right\}}{\frac{n + 1}{4}\left\{ 2a + \left( n - 1 \right)d \right\}}\]
\[ = \frac{2n}{n + 1}\]
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