Question

If S_n denotes the sum of first n terms of an A.P., prove that S₁₂ = 3(S₈ – S₄).

Answer 1

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

We know that, sum of first n terms = S_n = \[\frac{n}{2}\][2a + (n − 1)d]

Now,

S₄ = \[\frac{4}{2}\][2a + (4 − 1)d]

= 2(2a + 3d)

= 4a + 6d              ....(1)

S₈ = \[\frac{8}{2}\] [2a + (8 − 1)d]

= 4(2a + 7d)

= 8a + 28d            ....(2) 

S₁₂ = \[\frac{12}{2}\] [2a + (12 − 1)d]

= 6(2a + 11d)

= 12a + 66d          ....(3)

On subtracting (1) from (2), we get

S₈ − S₄ = 8a + 28d − (4a + 6d)

= 4a + 22d

Multiplying both sides by 3, we get

3(S₈ − S₄) = 3(4a + 22d)

= 12a + 66d

= S₁₂                 [From (3)]

Thus, S₁₂ = 3(S₈ − S₄).