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).