Let the sum of the first n terms of a non-constant A.P. a1,a2,a3,…….. be 50n+n(n−7)2A where A is a constant. If d is the common difference of this A.P., then the ordered pair d,a50 is equal to

A
$(A, 50 + 45 A)$
B
$(A, 55 + 45 A)$
C
$(50, 50 + 46 A)$
D
$(50, 50 + 45 A)$

Solution:

Given, $S_{n} = 50 n + \frac{n (n - 7) A}{2}$

We know that $a_{n} = S_{n} - S_{n - 1}$

$\begin{array}{l} = 50 n + 1 \frac{n (n - 7) A}{2} - 50 (n - 1) - \frac{(n - 1) (n - 8) A}{2} \\ = 50 + \frac{A}{2} [n^{2} - 7 n - n^{2} + 9 n - 8] = 50 + A (n - 4) \end{array}$

Now, $d = a_{n} - a_{n - 1} = 50 + A (n - 4) - 50 - A (n - 5) = A$

And $a_{50} = 50 + (50 - 4) A = 50 + 46 A$

Hence, ordered pair $(d, a_{50}) = (A, 50 + 46 A)$

Let the sum of the first $n$ terms of a non-constant A.P. $a_{1}, a_{2}, a_{3}, \dots \dots . .$ be $50 n + \frac{n (n - 7)}{2} A$ 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

Solution:

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