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Question
Find the sum of the following arithmetic progressions:
`(x - y)/(x + y),(3x - 2y)/(x + y), (5x - 3y)/(x + y)`, .....to n terms
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Solution
`(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))`
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