Question

Find the sum of the following arithmetic progressions:

`(x - y)^2, (x^2 + y^2), (x + y)^2` ...., to n terms

Answer 1

`(x - y)^2, (x^2 + y^2), (x + y)^2`...., to n terms

Common difference of the A.P. (d) = `a_2 - a_1`

`= (x^2  + y^2) - (x - y)^2`

`= x^2 + y^2 - (x^2 + y^2 - 2xy)`

`= x^2 + y^2 - x^2 - y^2 + 2xy`

= 2xy

Number of terms (n) = n

First term for the given A.P. `(a) = (x - y)^2`

So, using the formula we get,

`S_n = n/2 [2(x - y)^2 + (n -1)2xy]`

Now, taking 2 common from both the terms inside the bracket we get,

`= (n/2)[(2)(x -y)^2 +(2)(n -1)xy]`

`= (n/2)(2)[(x - y)^2 + (n -1)xy]`

`= (n)[(x - y)^2 + (n -1)xy]`

Therefore, the sum of first n terms for the given A.P. is `n[(x - y)^2 + (n -1)xy]`