Prove the following
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 :

  1. (1) $(2490,249)$

  2. (2) $(2480,249)$

  3. (3) $(2480,248)$

  4. (4) $(2490,248)$


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 $d=13$

$a_{2 j}-1+19 \times 13=248$

$s_{20}=\frac{20}{2}(2+19 \times 13)=2490$

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