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).