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)