Find the sum of the following arithmetic progression. x - y x + y , 3 x - 2 y x + y , 5 x - 3 y x + y , … . (to n terms)

Given A.P. is x - y x + y , 3 x - 2 y x + y , 5 x - 3 y x + y , … . (to n terms) Number of terms = n First term for the given A.P., a = x - y x + y Common difference of the A.P., d = a 2 - a 1 = 3 x - 2 y x + y - x - y x + y = 3 x - 2 y - ( x - y ) x + y = 3 x - 2 y - x + y x + y = 2 x - y x + y So, using the formula S n = n 2 [ 2 a + n - 1 d ] we get, S n = n 2 2 x - y x + y + n - 1 2 x - y x + y = n 2 2 x - 2 y x + y + ( n - 1 ( 2 x - y ) x + y = n 2 2 x - 2 y x + y + n 2 x - y - 1 ( 2 x - y ) x + y = n 2 2 x - 2 y x + y + n 2 x - y - 2 x + y x + y Now, on further solving the above equation we get, = n 2 2 x - 2 y + n 2 x - y - 2 x + y x + y = n 2 n 2 x - y - y x + y Therefore, the sum of first n terms for the given A.P. is n 2 n 2 x - y - y x + y .

Find the sum of the last ten terms of the A.P 9 , 10 , 12 , 14 , … … , 126 .

The given A.P is 8 , 10 , 12 , 14 , … . . , 126 . a = 8 and d = 2 . When this A.P. is reversed, we get, 126 , 124 , 122 , 120 , … . . So, first term becomes 126 and common difference becomes - 2 . The sum of last 10 terms of the given A.P. = The sum of the first 10 terms of the changed A.P S 10 = 10 2 2 × 126 + ( 10 - 1 ) ( - 2 ) = 5 [ 252 - 18 ] = 5 234 = 1170 Therefore, the sum of last ten terms of an A.P is 1170 .