Let $a_1,a_2,a_3,\mathrm{.......}$  be terms of an A.P. If $\frac{a_1+a_2+\mathrm{....}{a}_p}{a_1+a_2+\mathrm{....}+a_q}=\frac{p^2}{q^2},p\ne q,$  then $\frac{a_6}{a_{21}}$  equals

$\frac{\frac{p}{2}\left[2a_1+\left(p-1\right)d\right]}{\frac{q}{2}\left[2a_1+\left(q-1\right)d\right]}\Rightarrow \frac{2a_1+\left(p-1\right)d}{2a_1+\left(q-1\right)d}=\frac{p}{q}$

$\frac{a_1+\left(\frac{p-1}{2}\right)d}{a_1+\left(\frac{q-1}{2}\right)d}=\frac{p}{q}$

For $\frac{a_6}{a_{21}},p=11,q=41$      $\to \frac{a_6}{a_{21}}=\frac{11}{41}$