Let the sum of the first n terms of a non-constant AP `a_(1), a_(2), a_(3)...."be " 50n + (n (n-7))/(2)A`, where A is a constant. If d is the common difference of this AP, then the ordered pair `(d, a_(50))` is equal to

Options

A. `(A, 50 + 46A)`
B. `(50, 50 + 45A)`
C. `(50, 50 + 46A)`
D. `(A, 50 + 45A)`

Answer

A

Explanation

Key Idea Use the formula of sum of first n terms of AP, i.e., `S_(n) = (n)/(2) [2a + (n-1)d]`
Given AP, is
`a_(1), a_(2), a_(3)`.... Having sum of first n-terms
`= (n)/(2) [2a_(1) + (n -1)d]` [where, d is the common difference of AP]
`= 50n + (n (n-7))/(2) A` (given)
`rArr (1)/(2) [2a_(1) + (n-1) d] = 50 + (n-7)/(2)A`
`rArr (1)/(2) [2a_(1) + nd -d] = (50 - (7)/(2)A) + (n)/(2)A`
`rArr (a_(1) - (d)/(2)) + (nd)/(2) = (50 - (7)/(2) A) + (n)/(2) A`
On comparing corresponding term, we get
`d = A and a_(1) = 50 - (7)/(2) A`
`rArr a_(1) - (A)/(2) = 50 - (7)/(2) A " " [ :' d = A]`
`rArr a_(1) = 50 - 3A`
so `a_(50) = a_(1) + 49d`
`= (50 - 3A) + 49A " " [ :' d = A]`
`= 50 + 46 A`
Therefore, `(d, a_(50)) = (A, 50 + 46A)`