(d) : Given a_1, a_2, a_3, \ldots \text { be terms of A.P. }\\\n&\begin{aligned}\n& \frac{a_1+a_2+\cdots+a_p}{a_1+a_2+\cdots+a_q}=\frac{p^2}{q^2} \Rightarrow \frac{\frac{p}{2}\left[2 a_1+(p-1) d\right]}{\frac{q}{2}\left[2 a_1+(q-1) d\right]}=\frac{p^2}{q^2} \\\n& \Rightarrow \frac{2 a_1+(p-1) d}{2 a_1+(q-1) d}=\frac{p}{q} \\\n& \Rightarrow\left[2 a_1+(p-1) d\right] q=p\left[2 a_1+(q-1) d\right] \\\n& \Rightarrow 2 a_1(q-p)=d[(q-1) p-(p-1) q] \\\n& \Rightarrow 2 a_1(q-p)=d(q-p) \Rightarrow 2 a_1=d \\\n& \Rightarrow \frac{a_6}{a_{21}}=\frac{a_1+5 d}{a_1+20 d}=\frac{a_1+10 a_1}{a_1+40 a_1} \Rightarrow \frac{a_6}{a_{21}}=\frac{11}{41}\n\end{aligned}