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Question
In the arithmetic progression whose common difference is non-zero, the sum of first 3 n terms is equal to the sum of next n terms. Then the ratio of the sum of the first 2 n terms to the next 2 nterms is
Options
1/5
2/3
3/4
none of these
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Solution
1/5
\[S_{3n} = S_{4n} - S_{3n} \]
\[ \Rightarrow 2 S_{3n} = S_{4n} \]
\[ \Rightarrow 2 \times \frac{3n}{2}\left\{ 2a + \left( 3n - 1 \right)d \right\} = \frac{4n}{2}\left\{ 2a + \left( 4n - 1 \right)d \right\}\]
\[ \Rightarrow 3\left\{ 2a + \left( 3n - 1 \right)d \right\} = 2\left\{ 2a + \left( 4n - 1 \right)d \right\}\]
\[ \Rightarrow 6a + 9nd - 3d = 4a + 8nd - 2d\]
\[ \Rightarrow 2a + nd - d = 0\]
\[ \Rightarrow 2a + \left( n - 1 \right)d = 0 . . . . \left( 1 \right)\]
Required ratio: \[\frac{S_{2n}}{S_{4n} - S_{2n}}\]
\[\frac{S_{2n}}{S_{4n} - S_{2n}} = \frac{\frac{2n}{2}\left\{ 2a + \left( 2n - 1 \right)d \right\}}{\frac{4n}{2}\left\{ 2a + \left( 4n - 1 \right)d \right\} - \frac{2n}{2}\left\{ 2a + \left( 2n - 1 \right)d \right\}}\]
\[ = \frac{n\left( nd \right)}{2n\left( 3nd \right) - n\left( nd \right)}\]
\[ = \frac{1}{6 - 1}\]
\[ = \frac{1}{5}\]
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