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Question
Find the sum of the following arithmetic progressions
`(x - y)^2,(x^2 + y^2), (x + y)^2,.... to n term`
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Solution
`(x - y)^2,(x^2 + y^2), (x + y)^2,.... to n term`
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]`
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