Question

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

Answer 1

We know

S_n=`n/2`[2a+(n−1)d]

⇒S₂₀=`20/2`[2a+(20−1)d] and S₁₀=`10/2`[2a+(10−1)d]

⇒S₂₀=10[2a+19d] and S₁₀=5[2a+9d]

⇒S₂₀=20a+190d and S₁₀=10a+45d

3(S₂₀−S₁₀)=3(20a+190d−10a−45d)

= 3(10a+145d)

= 15(2a+29d)

=`30/2`[2a+(30−1)d]

= S₃₀

∴ S₃₀=3(S₂₀−S₁₀₎