Let a_1, a_2, a_3, ldots, a_{100} be an arith | vedclass

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(C) Let the common difference of the arithmetic progression be $d$. The sum of the first $p$ terms is given by $S_p = \frac{p}{2} [2a_1 + (p-1)d]$.Giv

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$9$

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(C) Let the common difference of the arithmetic progression be $d$. The sum of the first $p$ terms is given by $S_p = \frac{p}{2} [2a_1 + (p-1)d]$.Given $a_1 = 3$,we have $S_p = \frac{p}{2} [6 + (p-1)d]$.We are given $m = 5n$. Thus,$\frac{S_m}{S_n} = \frac{S_{5n}}{S_n} = \frac{\frac{5n}{2} [6 + (5n-1)d]}{\frac{n}{2} [6 + (n-1)d]} = 5 	imes \frac{6 - d + 5nd}{6 - d + nd}$.For this ratio to be independent of $n$,the coefficients of $n$ in the numerator and denominator must be proportional,or the expression must simplify to a constant.Let $f(n) = \frac{5nd + (6-d)}{nd + (6-d)}$. For this to be constant,we require $\frac{5d}{d} = \frac{6-d}{6-d}$,which is always true if $d 
eq 0$. However,for the ratio to be independent of $n$,we need the ratio of the coefficients of $n$ to be equal to the ratio of the constant terms: $\frac{5d}{d} = \frac{6-d}{6-d}$. This implies $5 = 1$,which is impossible unless the constant term is zero.If $6-d = 0$,then $d = 6$.Then $S_p = \frac{p}{2} [6 + (p-1)6] = \frac{p}{2} [6p] = 3p^2$.Then $\frac{S_{5n}}{S_n} = \frac{3(5n)^2}{3n^2} = \frac{75n^2}{3n^2} = 25$,which is independent of $n$.Thus,$d = 6$.Since $a_2 = a_1 + d = 3 + 6 = 9$.

Let $a_1, a_2, a_3, \ldots, a_{100}$ be an arithmetic progression with $a_1=3$ and $S_p=\sum_{i=1}^p a_i, 1 \leq p \leq 100$. For any integer $n$ with $1 \leq n \leq 20$,let $m=5n$. If $\frac{S_m}{S_n}$ does not depend on $n$,then $a_2$ is

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