Question

The sum of n terms of three arithmetical progression are S₁ , S₂ and S₃ . The first term of each is unity and the common differences are 1, 2 and 3 respectively. Prove that S₁ + S₃ = 2S₂

Answer 1

We have,

S₁ = Sum of n terms of an A.P. with first term 1 and common difference 1

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

S₂ = Sum of n terms of an A.P. with first term 1 and common difference 2

`= \frac { n }{ 2 } [2 × 1 + (n – 1) × 2] = n^2`

S₃ = Sum of n terms of an A.P. with first term 1 and common difference 3

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

Now,

`S_1 + S_3 = \frac { n }{ 2 } (n + 1) + \frac { n }{ 2 } (3n – 1)`

= 2n² and S₂ = n²

Hence S₁ + S₃ = 2S₂