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

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Solution

Formula,
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{10} = \frac{10}{2} [2 a + (10 - 1) d] = 5 [2 a + 9 d]$
$S_{20} = \frac{20}{2} [2 a + (20 - 1) d] = 10 [2 a + 19 d]$
$S_{30} = \frac{30}{2} [2 a + (30 - 1) d] = 15 [2 a + 29 d]$
$S_{20} - S_{10} = 10 [2 a + 19 d] - 5 [2 a + 9 d]$
$= 10 a + 145 d$
$= 5 [2 a + 29 d]$

We have,

$S_{30} = 15 [2 a + 29 d] = 3 \times {5 [2 a + 29 d]}$

$∴ S_{30} = 3 \times (S_{20} - S_{10})$

Hence, proved.

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