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

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Solution

If $S_{n}$ = Sum of first $n$ terms of $A P$
$S_{30} = 3 (S_{20} - S_{10})$ ........ $(1)$
On solving $L H S$
We know $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{30} = \frac{n}{2} [2 a + (n - 1) d] = 15 [2 a + 29 d]$ ........ $(2)$
On solving $R H S$
$3 (S_{20} - S_{10}) = 3 (\frac{2 a}{2} [2 a + 19 d] - \frac{10}{2} [2 a + 9 d])$
$= 3 [20 a + 190 d - 10 a - 45 d]$
$= 3 [10 a + 145 d]$
$= 30 a + 435 d$
$3 (S_{20} - S_{10}) = 15 [2 a + 29 d]$
Hence $[L H S = R H S] = 15 [2 a + 29 d]$
Hence proved

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