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Question
Mark the correct alternative in the following question:
Let Sn denote the sum of first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to
Options
4
6
8
10
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Solution
\[As, S_{2n} = 3 S_n \]
\[ \Rightarrow \frac{2n}{2}\left[ 2a + \left( 2n - 1 \right)d \right] = \frac{3n}{2}\left[ 2a + \left( n - 1 \right)d \right]\]
\[ \Rightarrow 2\left[ 2a + \left( 2n - 1 \right)d \right] = 3\left[ 2a + \left( n - 1 \right)d \right]\]
\[ \Rightarrow 4a + 2\left( 2n - 1 \right)d = 6a + 3\left( n - 1 \right)d\]
\[ \Rightarrow 4a + 4nd - 2d = 6a + 3nd - 3d\]
\[ \Rightarrow 6a - 4a + 3nd - 3d - 4nd + 2d = 0\]
\[ \Rightarrow 2a - nd - d = 0\]
\[ \Rightarrow 2a - \left( n + 1 \right)d = 0\]
\[ \Rightarrow 2a = \left( n + 1 \right)d . . . . . \left( i \right)\]
\[\text { Now }, \]
\[\frac{S_{3n}}{S_n} = \frac{\frac{3n}{2}\left[ 2a + \left( 3n - 1 \right)d \right]}{\frac{n}{2}\left[ 2a + \left( n - 1 \right)d \right]}\]
\[ = \frac{3\left[ \left( n + 1 \right)d + \left( 3n - 1 \right)d \right]}{\left[ \left( n + 1 \right)d + \left( n - 1 \right)d \right]} \left[ \text { Using } \left( i \right) \right]\]
\[ = \frac{3\left[ nd + d + 3nd - d \right]}{\left[ nd + d + nd - d \right]}\]
\[ = \frac{3\left[ 4nd \right]}{\left[ 2nd \right]}\]
\[ = 3 \times 2\]
\[ = 6\]
Hence, the correct alternative is option (b).
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