Answer 1

`(x - y)/(x + y),(3x - 2y)/(x + y), (5x - 3y)/(x + y)`,  .....to n terms

Number of terms (n) = n

Number of terms (n) = n = `((x - y)/(x + y))`

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

`= ((3x - 2)/(x + y)) - (x - y)/(x + y)`

`= ((3x - 2y) - (x - y))/(x +y)`

`= (3x - 2y - x + y)/(x + y)`

`= (2x - y)/(x - y)`

So using the formula we get

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

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

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

Now, on further solving the above equation we get,

`= (n/2)((2x - 2y + n(2x - y) - 2x + y)/(x + y))`

`= (n/2) ((n(2x - y) - y)/(x + y))`

Therefore, the sum of first n terms for the given A.P. is `(n/2) ((n(2x - y) - y)/(x + y))`