Let the sum of the first n terms of a non-con | vedclass

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(A) Given the sum of the first $n$ terms: $S_n = 50n + \frac{n(n - 7)}{2}A$.The $n^{th}$ term $T_n$ is given by $T_n = S_n - S_{n-1}$.$T_n = 50n + \fr

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$(A, 50 + 46A)$

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(A) Given the sum of the first $n$ terms: $S_n = 50n + \frac{n(n - 7)}{2}A$.The $n^{th}$ term $T_n$ is given by $T_n = S_n - S_{n-1}$.$T_n = 50n + \frac{n(n - 7)}{2}A - [50(n - 1) + \frac{(n - 1)(n - 8)}{2}A]$.$T_n = 50 + \frac{A}{2} [n^2 - 7n - (n^2 - 9n + 8)]$.$T_n = 50 + \frac{A}{2} [2n - 8] = 50 + A(n - 4)$.The common difference $d = T_n - T_{n-1} = [50 + A(n - 4)] - [50 + A(n - 5)] = A$.To find $a_{50}$,substitute $n = 50$ into the expression for $T_n$:$a_{50} = 50 + A(50 - 4) = 50 + 46A$.Thus,the ordered pair $(d, a_{50})$ is $(A, 50 + 46A)$.

Let the sum of the first $n$ terms of a non-constant $A.P.$,$a_1, a_2, a_3, \dots$ be $S_n = 50n + \frac{n(n - 7)}{2}A$,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

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