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: , 3
Solution:
$a_{n}=a_{1}+(n-1) d$
$\Rightarrow 300=1+(\mathrm{n}-1) \mathrm{d}$
$\Rightarrow(\mathrm{n}-1) \mathrm{d}=299=13 \times 23$
since, $\mathrm{n} \in[15,50]$
$\therefore \mathrm{n}=24$ and $\mathrm{d}=13$
$a_{n-4}=a_{20}=1+19 \times 13=248$
$\Rightarrow a_{n-4}=248$
$S_{n-4}=\frac{20}{2}\{1+248\}=2490$
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