Question

If Sn denotes the sum of first n terms of an AP then 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

We know that sum of n terms of an AP sn = (n/2)[2a + (n - 1) * d]

(i) S12 = (12/2)[2a + (12 - 1) * d]

         = 6[2a + 11d]

         = 12a + 66d

(ii) S8 = 8/2[2a + (8 - 1) * d]

          = 4[2a + 7d]

          = 8a + 28d

(iii) S4 = (4/2)[2a + (4 - 1) * d]

          = 2[2a + 3d]

          = 4a + 6d.

From (ii) & (iii)

⇒ S8 - S4 = 8a + 28d - 4a - 6d

                = 4a + 22d.

               

Now,

⇒ 3(S8 - S4) = 3(4a + 22d)

                    = 12a + 66d

                    = S12.

RHS = LHS.

Hope it helps!