JEE PYQ: Sequence And Series Question 28
Question 28 - 2020 (04 Sep Shift 2)
Let $a_1, a_2, \ldots, a_n$ be a given A.P. whose common difference is an integer and $S_n = a_1 + a_2 + \ldots + a_n$. If $a_1 = 1$, $a_n = 300$ and $15 \leq n \leq 50$, then the ordered pair $(S_{n-4}, a_{n-4})$ is equal to:
(1) $(2490, 249)$
(2) $(2480, 249)$
(3) $(2480, 248)$
(4) $(2490, 248)$
Show Answer
Answer: (4)
Solution
Given that $a_1 = 1$ and $a_n = 300$ and $d \in Z$.
$\therefore 300 = 1 + (n-1)d$
$\Rightarrow d = \frac{299}{(n-1)} = \frac{23 \times 13}{(n-1)}$
$\therefore d$ is an integer
$\therefore n - 1 = 13$ or $23$
$\Rightarrow n = 14$ or $24$ ($\because 15 \leq n \leq 50$)
$\Rightarrow n = 24$ and $d = 13$
$a_{20} = 1 + 19 \times 13 = 248$
$s_{20} = \frac{20}{2}(2 + 19 \times 13) = 2490$
BETA
