Question

if Sn denotes first n termn of an AP, prove that S12= 3(S8-S4)

Answer 1

S(n)= n/2(2a+(n-1)d)
so,
S(12)=12/2(2a+(12-1)d)
=6(2a+11d)
=3(4a+22d)

Again, 3(S(8)-S(4))= 3(8/2(2a+(8-1)d) - 4/2(2a+(4-1)d)) = 3(4(2a+7d) - 2(2a+3d))
= 3(8a+28d-(4a+6d))
= 3(8a+28d-4a-6d)
= 3(4a+22d)

therefore, S(12)=3(S(8)-S(4))

Answer 2

$\Large \bf Solution :$

Let the first term be a and common difference be d.

We have to prove : $\sf S_{12} = 3(S_{8} - S_{4})$

$\underline {\bf In \: RHS, \: we \: have :}$$\sf 3(S_{8} - S_{4})$

Now,

$\underline{\boxed{\sf S_{n} = \dfrac{n}{2} . [2a+(n-1)d]}}$

$\sf : \implies 3 \Bigg[\dfrac{8}{2}(2a+(8-1)d) - \dfrac{4}{2}(2a+(4-1)d)\Bigg]$

$\sf : \implies 3 \Bigg[\cancel{\dfrac{8}{2}}(2a+7d) - \cancel{\dfrac{4}{2}}(2a+3d)\Bigg]$

$\sf : \implies 3 [4(2a+7d) - 2(2a+3d)]$

$\sf : \implies 3 \times 2[2(2a+7d) - (2a+3d)]$

$\sf : \implies 6(4a+14d - 2a+3d)$

$\sf : \implies 6(2a+11d)$

$\underline {\bf In \: LHS, \: we \: have :}$$\sf S_{12}$

Now,

$\underline{\boxed{\sf S_{n} = \dfrac{n}{2} . [2a+(n-1)d]}}$

$\sf : \implies \dfrac{12}{2} (2a + (12-1)d)$

$\sf : \implies \cancel{\dfrac{12}{2}} (2a + 11d)$

$\sf : \implies 6 (2a + 11d)$

$\bf As, \: LHS = RHS,$

$\underline{\underline{\bf Hence, \: Proved.}}$