# Let a1, a2 , ................an be given A.P, whose common difference is an integer and $S_{n}=a_{1}+a_{2}+..........a_{n}$. 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: Option: 1 (2490, 249) Option: 2 (2480, 249) Option: 3 (2480, 248) Option: 4 (2490, 248)

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

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