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Sum of n Terms in AP

Let the sum o...

Question

Let the sum of the first n terms of a non-constant A.P., $a_{1}, a_{2}, a_{3}, \dots$ 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 orderd pair $(d, a_{50}$ ) is equal to:

(50, 50 + 46 A)

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(A, 50 + 46 A)

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(A, 50 + 45 A)

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(50, 50 + 45 A)

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Solution

The correct option is B $(A, 50 + 46 A)$
$S_{n} = 50 n + \frac{n (n - 7)}{2} A$
and we know that
$n t h term, T_{n} = S_{n} - S_{n - 1} = 50 n + \frac{n (n - 7)}{2} A - 50 (n - 1) - \frac{(n - 1) (n - 8)}{2} A = 50 + \frac{A}{2} [n^{2} - 7 n - n^{2} + 9 n - 8] = 50 + A (n - 4)$
$and d = T_{n} - T_{n - 1} = 50 + A (n - 4) - 50 - A (n - 5) = A ∴ T_{50} = 50 + A (50 - 4) ∴ (d, a_{50}) = (A, 50 + 46 A)$

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