The sum of $n, 2 n$ and $3 n$ terms of an $A P$ are $x, y, z$ . Prove that $z = 3 (y - x)$ .

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Solution

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

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

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

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

$∴ 3 (y - x)$

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

$= 3 \frac{n}{2} [2 a + (4 n - 2 - n + 1) d]$

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

$= z$

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