Question

If Sn demotes the sum of first n terms of an AP. Prove that S12=3(S8-S4)​

Answer 1

Answer:-

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 (Proved)