Let a_{1}, a_{2}, ldots, a_{n} be a given A.P | vedclass

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(C) The $n$-th term of an $A.P.$ is given by $a_{n} = a_{1} + (n-1)d$.Given $a_{1} = 1$ and $a_{n} = 300$,we have $300 = 1 + (n-1)d$,which implies $(n

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$(2490, 248)$

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(C) The $n$-th term of an $A.P.$ is given by $a_{n} = a_{1} + (n-1)d$.Given $a_{1} = 1$ and $a_{n} = 300$,we have $300 = 1 + (n-1)d$,which implies $(n-1)d = 299$.The prime factorization of $299$ is $13 	imes 23$.Since $15 \leq n \leq 50$,we have $14 \leq n-1 \leq 49$.The possible values for $(n-1)$ are $23$ or $13$ (not possible as $n-1 \geq 14$).Thus,$n-1 = 23 \Rightarrow n = 24$ and $d = 13$.We need to find $(S_{n-4}, a_{n-4}) = (S_{20}, a_{20})$.$a_{20} = a_{1} + 19d = 1 + 19(13) = 1 + 247 = 248$.$S_{20} = \frac{20}{2}(a_{1} + a_{20}) = 10(1 + 248) = 10(249) = 2490$.Therefore,the ordered pair is $(2490, 248)$.

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 $(S_{n-4}, a_{n-4})$ is equal to

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