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

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Solution

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

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

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

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

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

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

Hence, $S_{3} = 2 (S_{2} - S_{1})$

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