Let a_1, a_2, a_3, ldots be terms of an A.P. | vedclass

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(D) The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} [2a_1 + (n-1)d]$.Given $\frac{S_p}{S_q} = \frac{p^2}{q^2}$,we have $\fr

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$\frac{11}{41}$

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(D) The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} [2a_1 + (n-1)d]$.Given $\frac{S_p}{S_q} = \frac{p^2}{q^2}$,we have $\frac{\frac{p}{2} [2a_1 + (p-1)d]}{\frac{q}{2} [2a_1 + (q-1)d]} = \frac{p^2}{q^2}$.Simplifying,we get $\frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p}{q}$.Dividing the numerator and denominator by $2$,we get $\frac{a_1 + (\frac{p-1}{2})d}{a_1 + (\frac{q-1}{2})d} = \frac{p}{q}$.To find $\frac{a_6}{a_{21}}$,we need the index $n$ such that $\frac{n-1}{2} = 5$ (for $a_6$) and $\frac{n-1}{2} = 20$ (for $a_{21}$).For $a_6$,$p-1 = 10 \Rightarrow p = 11$.For $a_{21}$,$q-1 = 40 \Rightarrow q = 41$.Substituting these values,$\frac{a_6}{a_{21}} = \frac{p}{q} = \frac{11}{41}$.

Let $a_1, a_2, a_3, \ldots$ be terms of an $A.P.$ If $\frac{a_1 + a_2 + \ldots + a_p}{a_1 + a_2 + \ldots + a_q} = \frac{p^2}{q^2}$ for $p \neq q$,then $\frac{a_6}{a_{21}}$ equals:

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