JEE PYQ: Sequence And Series Question 48
Question 48 - 2019 (09 Apr Shift 1)
Let the sum of the first $n$ terms of a non-constant A.P. $a_1, a_2, a_3, \ldots$ 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:
(1) $(50, 50 + 46A)$
(2) $(50, 50 + 45A)$
(3) $(A, 50 + 45A)$
(4) $(A, 50 + 46A)$
Show Answer
Answer: (4)
Solution
$\therefore d = a_2 - a_1 = (S_{n,2} - S_{n,1}) - S_1$
$\Rightarrow d = A$
$a_{50} = a_1 + 49 \times d$
$= (50 - 3A) + 49A = 50 + 46A$
So, $(d, a_{50}) = (A, 50 + 46A)$
BETA
