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  4. Let the sum of the first n terms of a non-constant A.P., a1, a2, a3, ......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, a50) is equal to

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Q. Let the sum of the first $n$ terms of a non-constant $A.P$., $a_1, a_2, a_3, ......$be 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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Solution:

$S_n = 50n + \frac{n(n-7)}{2}$A
$T_n = S_n - S_{n-1}$
$= 50 n + \frac{n(n-7)}{2}A-50(n-1) - \frac{(n-1(n-8)}{2}A$
$= 50 + \frac{A}{2} [n^2 - 7n -n^2 + 9n - 8]$
= 50 + A(n - 4)
$d = T_n - T_{n-1}$
= $50 + A(n - 4) - 50 - A(n - 5)$
= $A$
$T_{50} = 50 + 46A$
$(d. A_{50}) = (A, 50 +46A)$