Question

If S_n denotes the sum of the first n terms of an A.P., prove that S₃₀ = 3(S₂₀ − S₁₀)

 

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,
S₁₀ = \[\frac{10}{2}\] [2a + (10 − 1)d]
      = 5(2a + 9d)
      = 10a + 45d          ....(1) 

S₂₀ = \[\frac{20}{2}\] [2a + (20 − 1)d]
      = 10(2a + 19d)
      = 20a + 190d        ....(2) 

S₃₀ = \[\frac{30}{2}\] [2a + (30 − 1)d]
      = 15(2a + 29d)
      = 30a + 435d        ....(3)

On subtracting (1) from (2), we get
S₂₀ − S₁₀ = 20a + 190d − (10a + 45d)
                = 10a + 145d

On multiplying both sides by 3, we get
3(S₂₀ − S₁₀) = 3(10a + 145d)
                    = 30a + 435d
                    = S₃₀                   [From (3)]

Hence, S₃₀ = 3(S₂₀ − S₁₀)