If $S_{n}$ denotes thesum of first n terms of an A.P., prove that $S_{12} = 3 (S_{8} - S_{4})$ .

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Solution

Let a is the first term of Ap and d is the common difference
$S_{n} = \frac{n}{2} (2 a + (n - 1) d)$

now $S_{12} = \frac{12}{2} (2 a + (12 - 1) d) = 12 a + 66 d$
$S_{8} = \frac{8}{2} (2 a + 7 d) = 8 a + 28 d$
$S4=frac42(2a+3d)=4a+6d$
LHS= $S_{12} = 12 a + 66 d$
RHS=3 $(S_{8} - S_{4}) = 3 (8 a + 28 d - 4 a - 6 d) = 12 a + 66 d$
LHS =RHS

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