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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 S 1 S 2 =

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Question

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 \[\frac{S_1}{S_2}\] = 

Options

  • \[\frac{2n}{n + 1}\]

  • \[\frac{n}{n + 1}\]

  • \[\frac{n + 1}{2n}\]

  • \[\frac{n + 1}{n}\]

MCQ
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Solution

\[\frac{2n}{n + 1}\]

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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Chapter 19: Arithmetic Progression - Exercise 19.9 [Page 52]

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RD Sharma Mathematics [English] Class 11
Chapter 19 Arithmetic Progression
Exercise 19.9 | Q 19 | Page 52

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