Let the sum of $n, 2 n, 3 n$ terms of an A.P. be $S_{1}, S_{2}$ and $S_{3}$ , respectively, show that $S_{3} = 3 (S_{2} - S_{1})$ .

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Solution

Let $a$ and $d$ be the first term and the common difference of the A.P. respectively.

Therefore,

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

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

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

From (1) and (2), we obtain

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

$= n {\frac{4 a + 4 n d - 2 d - 2 a - n d + d}{2}}$

$= n [\frac{2 a + 3 n d - d}{2}]$

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

$∴ 3 (S_{2} - S_{1}) = \frac{3 n}{2} [2 a + (3 n - 1) d] = S_{3}$ $[F r o m (3)]$ $[h e n c e p r o v e d]$

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