if $S_{n}$ denotes sum of the n terms of AP than prove that $S_{12} = 3 \times (S_{8} - S_{4})$

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Solution

Given:
$P . T S_{12} = 3 \times (S_{8} - S_{4})$

Let first term $= a$ and the common difference $= d$

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

$S_{12} = \frac{12}{2} [2 a + 11 d] = 6 [2 a + 11 d]$ ............. $(1)$

Now, $S_{8} - S_{4} = \frac{8}{2} [2 a + 7 d] - \frac{4}{2} [2 a + 3 d]$

$= 4 [2 a + 7 d] - 2 [2 a + 3 d]$

$(8 a - 4 a) + (28 d - 6 d)$

$4 a + 22 d = 2 (2 a + 11 d)$ ...... $(2)$

By $(1)$ & $(2)$

$8 a + 4 d = \frac{6 (2 a + 11 d)}{3} \Rightarrow 3 (8 a + 4 d) = 6 (2 a + 11 d) \Rightarrow 3 (S_{8} - S_{4}) = S_{12}$

Hence $S_{12} = 3 \times (S_{8} - S_{4})$

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