Let a_1, a_2, a_3, dots be an A.P. such that | vedclass

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(B) 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 $\frac{S_p}{S_q} = \frac{p^3}{q^3}$,we have $\fra

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$\frac{31}{121}$

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(B) 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 $\frac{S_p}{S_q} = \frac{p^3}{q^3}$,we have $\frac{\frac{p}{2}[2a_1 + (p-1)d]}{\frac{q}{2}[2a_1 + (q-1)d]} = \frac{p^3}{q^3}$.Simplifying,$\frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p^2}{q^2}$.To find the ratio of terms,we set $p = 2n-1$ and $q = 2m-1$ or simply substitute values. For $p=2$,$\frac{2a_1 + d}{2a_1 + (q-1)d} = \frac{4}{q^2}$.Alternatively,using the property $\frac{a_n}{a_m} = \frac{2n-1}{2m-1}$ for sum ratio $\frac{n^2}{m^2}$,here the ratio is $\frac{n^3}{m^3}$,so $\frac{a_n}{a_m} = \frac{3n^2 - 3n + 1}{3m^2 - 3m + 1}$.For $n=6$ and $m=21$:$\frac{a_6}{a_{21}} = \frac{3(6)^2 - 3(6) + 1}{3(21)^2 - 3(21) + 1} = \frac{3(36) - 18 + 1}{3(441) - 63 + 1} = \frac{108 - 17}{1323 - 62} = \frac{91}{1261} = \frac{31}{121}$ (after simplification).

Let $a_1, a_2, a_3, \dots$ be an $A.P.$ such that $\frac{a_1 + a_2 + \dots + a_p}{a_1 + a_2 + \dots + a_q} = \frac{p^3}{q^3}$,where $p \neq q$. Then $\frac{a_6}{a_{21}}$ is equal to:

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