Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3 (S2– S1)

Let’s take a and d to be the first term and the common difference of the A.P. respectively.

So, we have

\( s_n=\dfrac{n}{2}[2a+(n-1)d]\)

Therefore,

\( s_1=\dfrac{n}{2}[2a+(n-1)d]\)

\( s_2=\dfrac{2n}{2}[2a+(n-1)d]\)

= \( n[2a+(n-1)d]\)

\( s_n=\dfrac{3n}{2}[2a+(n-1)d]\)

Subtract S₁ from S₂ .

\( S_2-S_1=S_2\)

=\(n [2a+(n-1)d]-\dfrac{n}{2}[2a+(n-1)d]\)

= \( n[\dfrac{4a+4nd-2d-2a-nd+d}{2}]\)

= \( n[\dfrac{2a+3nd-d}{2}]\)

= \( \dfrac{n}{2}[2a+(3n-1)d]\)

Therefore,

S₃= \( \dfrac{3n}{2}[2a+(3n-1)d]\)

= 3(S₂-S₁)

Therefore, S₃= 3(S₂-S₁) is proved