Sum

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

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#### Solution

We have,

S_{1} = 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]`

S_{2} = 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`

S_{3} = 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)`

= 2n^{2} and S_{2} = n^{2}

Hence S_{1} + S_{3} = 2S_{2}

Concept: Sum of First n Terms of an AP

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