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)