Question:
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 :
Correct Option: , 4
Solution:
Given that $a_{1}=1$ and $a_{n}=300$ and $d \in \mathbf{Z}$
$\therefore 300=1+(n-1) d$
$\Rightarrow d=\frac{299}{(n-1)}=\frac{23 \times 13}{(n-1)}$
$\because d$ is an integer
$\therefore n-1=13$ or 23
$\Rightarrow n=14$ or $24 \quad(\because 15 \leq n \leq 50)$
$\Rightarrow n=24$ and $d=13$
$a_{2 j}-1+19 \times 13=248$
$s_{20}=\frac{20}{2}(2+19 \times 13)=2490$