If sum of n, 2n, 3n terms of an A.P. are $S_{1}, S_{2}, S_{3}$ respectively, then the value of $\frac{S_{3}}{S_{2} - S_{1}}$ is

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Solution

The correct option is A
3

$S_{1} = \frac{n}{2} [2 a + (n - 1) d] S_{2} = \frac{2 n}{2} [2 a + (2 n - 1) d] S_{3} = \frac{3 n}{2} [2 a + (3 n - 1) d] S_{2} - S_{1} = \frac{n}{2} [2 a + (3 n - 1) d] 3 (S_{2} - S_{1}) = S_{3} \Rightarrow \frac{S_{3}}{S_{2} - S_{2}} = 3$

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The sum of n, 2n, 3n terms of an A.P. are $S_{1}$ , $S_{2}$ , $S_{3}$ respectively. Prove that $S_{3}$ = 3( $S_{2}$ – $S_{1}$ ).

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