a_1, a_2, a_3, dots, a_{100} are in an arithm | vedclass

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(B) Let the common difference be $d$. The sum of the first $p$ terms is $S_p = \frac{p}{2}[2a_1 + (p-1)d]$.Given $m = 5n$,we have $\frac{S_m}{S_n} = \

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$9$

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(B) Let the common difference be $d$. The sum of the first $p$ terms is $S_p = \frac{p}{2}[2a_1 + (p-1)d]$.Given $m = 5n$,we have $\frac{S_m}{S_n} = \frac{\frac{5n}{2}[2a_1 + (5n-1)d]}{\frac{n}{2}[2a_1 + (n-1)d]} = 5 	imes \frac{2a_1 - d + 5nd}{2a_1 - d + nd}$.For the ratio to be independent of $n$,the coefficients of $n$ must be proportional,or the term $(2a_1 - d)$ must be $0$.Setting $2a_1 - d = 0$,we get $d = 2a_1$.Given $a_1 = 3$,we have $d = 2(3) = 6$.Thus,$a_2 = a_1 + d = 3 + 6 = 9$.

$a_1, a_2, a_3, \dots, a_{100}$ are in an arithmetic progression,where $a_1 = 3$ and $S_p = \sum_{i=1}^p a_i, 1 \le p \le 100$. For any integer $n$,let $m = 5n$. If $S_m/S_n$ is independent of $n$,then $a_2 = \dots$

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