Question

if sn denotes the sum of first n terms of an ap, prove that s12=3(s8-s4)​

Answer 1

let a is the first term of Ap and d is the common difference

Sn=n/2 {2a+(n-1) d

now S12=12/2 {2a+(12-1) d}=12a+66d

S8=8/2 {2a+7d}=8a+28d

S4=4/2 {2a+3d}=4a+6d

LHS=S12=12a+66d

RHS=3 (S8-S4)=3 (8a+28d-4a-6d)=12a+66d

LHS =RHS

Answer 2

Solution :

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

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

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

Now,

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

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

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

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

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

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

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

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

Now,

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

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

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

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

As, LHS = RHS,

Hence, Proved.