Question

If the sum of first n, 2n and 3n terms of an AP be S₁, S₂ and S₃ respectively, then prove that S₃ = 3 (S₂ − S₁).

Answer 1

Let a be the first term and d the common difference of the given AP. Then,
S₁ = sum of first n terms of the given AP
S₂ = sum of first 2n terms of the given AP
S₃ = sum of first 3n terms of the given AP
Therefore,
$S_1=\frac{n}{2}\left\{2a+\left(n-1\right)d\right\}$
$S_2=\frac{2n}{2}\left\{2a+\left(2n-1\right)d\right\}$
$S_3=\frac{3n}{2}\left\{2a+\left(3n-1\right)d\right\}$
Thus,
$3\left(S_2-S_1\right)=3\left[\left\{2na+n\left(2n-1\right)d\right\}-\left\{na+\frac{1}{2}n\left(n-1\right)d\right\}\right]$
$=3\left[na+\frac{3}{2}{n}^{2}d-\frac{1}{2}nd\right]=\frac{3n}{2}\left[2a+3nd-d\right]$
$=\frac{3n}{2}\left[2a+\left(3n-1\right)d\right]=S_3$
Hence, S₃ = 3 (S₂ − S₁)