Question

The sum of n, 2n, 3n terms of an A.P. are S₁ , S₂ , S₃ respectively. Prove that S₃ = 3(S₂ – S₁ )

Answer 1

Let a be the first term and d be the common difference of the given A.P. Then,

S₁ = Sum of n terms

`⇒ S_1 = \frac { n }{ 2 } {2a + (n – 1)d} ….(i)`

S_2 = Sum of 2n terms

`⇒ S_2 = \frac { 2n }{ 2 } [2a + (2n – 1) d] ….(ii)`

and, S₃ = Sum of 3n terms

`⇒ S_3 = \frac { 3n }{ 2 } [2a + (3n – 1) d] ….(iii)`

Now, S₂ – S₁

`= \frac { 2n }{ 2 } [2a + (2n – 1) d] – \frac { n }{ 2 } [2a + (n –1) d]`

`S_2 – S_1 = \frac { n }{ 2 } [2 {2a + (2n – 1)d} – {2a + (n – 1)d}]`

`= \frac { n }{ 2 } [2a + (3n – 1) d]`

`\∴ 3(S_2 – S_1 ) = \frac { 3n }{ 2 } [2a + (3n – 1) d] = S_3 ` [Using (iii)]

Hence, S₃ = 3 (S₂ – S₁ )