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

Let the sum of the first n terms of a non-constant A.P., 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 A.P., then the ordered pair (d, a_(50)) is equal to

Let the sum of the first n terms of a non-constant A.P., 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 A.P., then the ordered pair (d, a_(50)) is equal to

The sum of first n terms of an AP is 3n^(2) + 4n . Find: a] a_(25) b] a_(10)

The sixth term of an A.P.,a_(1),a_(2),a_(3),.........,a_(n) is 2. If the quantity a_(1)a_(4)a_(5), is minimum then then the common difference of the A.P.

If a_(1), a_(2), a_(3) ,... are in AP such that a_(1) + a_(7) + a_(16) = 40 , then the sum of the first 15 terms of this AP is

If a_(1), a_(2), a_(3) ,... are in AP such that a_(1) + a_(7) + a_(16) = 40 , then the sum of the first 15 terms of this AP is

If a_(1), a_(2), a_(3) ,... are in AP such that a_(1) + a_(7) + a_(16) = 40 , then the sum of the first 15 terms of this AP is

Let a_(n) be the nth term of an AP, if sum_(r=1)^(100)a_(2r)= alpha " and "sum_(r=1)^(100)a_(2r-1)=beta , then the common difference of the AP is

Let a_(n) be the nth term of an AP, if sum_(r=1)^(100)a_(2r)= alpha " and "sum_(r=1)^(100)a_(2r-1)=beta , then the common difference of the AP is

Let a_(n) be the nth term of an AP, if sum_(r=1)^(100)a_(2r)= alpha " and "sum_(r=1)^(100)a_(2r-1)=beta , then the common difference of the AP is