Let the sum of the first n terms of a non-constant A.P.

Question

Let the sum of the first $n$ terms of a non-constant A.P. $a_1,a_2,a_3,\ldots$ be $50n+\dfrac{n(n-7)}{2}A$, where $A$ is a constant. If $d$ is the common difference, then the ordered pair $(d, a_{50})$ is equal to:

Accepted Answer

$S_n = \left(50-\dfrac{7A}{2}\right)n+\dfrac{A}{2}n^2$, so $a_1=S_1=50-3A$, and $d=2\cdot\dfrac{A}{2}=A$.

$a_{50} = a_1+49d = (50-3A)+49A = 50+46A.$

$(d,a_{50}) = (A, 50+46A).$

Solutions · 1