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8. Sequences and Series
hard
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 $\left( S _{ n -4}, a _{ n -4}\right)$ is equal to
A
$(2480,249)$
B
$(2490,249)$
C
$(2490,248)$
D
$(2480,248)$
(JEE MAIN-2020)
Solution
$\quad a_{n}=a_{1}+(n-1) d$
$\Rightarrow 300=1+(n-1) d$
$\Rightarrow \quad(n-1) d=299=13 \times 23$
since, $n \in[15,50]$
$\therefore n=24$ and $d=13$
$a_{n-4}=a_{20}=1+19 \times 13=248$
$\Rightarrow a_{n-4}=248$
$S_{n-4}=\frac{20}{2}\{1+248\}=2490$
Standard 11
Mathematics
