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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,$\frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p^2}{q^2}$.Cross-multiplying,$q^2(2a_1 + (p-1)d) = p^2(2a_1 + (q-1)d)$.$2a_1q^2 + p q^2 d - q^2 d = 2a_1p^2 + p^2 q d - p^2 d$.$2a_1(q^2 - p^2) = d(p^2 q - p q^2 + q^2 - p^2)$.$2a_1(q-p)(q+p) = d(pq(p-q) - (p-q)(p+q))$.$2a_1(q-p)(q+p) = d(p-q)(pq - p - q)$.Since $p 
eq q$,we divide by $(q-p)$: $2a_1(q+p) = -d(pq - p - q)$.For this to hold for all $p, q$,we compare coefficients or substitute values. Using $p=1, q=2$: $\frac{a_1}{a_1+a_2} = \frac{1}{8}$ $\Rightarrow 8a_1 = a_1 + a_1 + d$ $\Rightarrow d = 6a_1$.Then $\frac{a_6}{a_{21}} = \frac{a_1 + 5d}{a_1 + 20d} = \frac{a_1 + 5(6a_1)}{a_1 + 20(6a_1)} = \frac{31a_1}{121a_1} = \frac{31}{121}$.

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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