If $S_{n}$ denotes the sum of first $n$ terms of an $A . P$ , prove that $S_{30} = 3 (S_{20} - S_{10})$

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Solution

$S_{30} = \frac{30}{2} (2 a + 29 d)$
$S_{30} = 15 (2 a + 29 d) = 30 a + 435 d$ ...... $(1)$
$S_{20} - S_{10} = \frac{20}{2} (2 a + 19 d) - \frac{10}{2} (2 a + 9 d)$
$= 10 (2 a + 9 d) - 5 (2 a + 9 d)$
$= 20 a + 190 d - 10 a - 45 d$
$= 10 a + 145 d$
$3 \times (S_{20} - S_{10}) = 3 (10 a + 145 d) = 30 a + 435 d$ ...... $(2)$
From $(1)$ and $(2)$
$S_{30} = 3 (S_{20} - S_{10})$

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