Question
Let $${a_1},{a_2},{a_3}......$$ be terms on A.P. If $$\frac{{{a_1} + {a_2} + ...... + {a_p}}}{{{a_1} + {a_2} + ...... + {a_q}}} = \frac{{{p^2}}}{{{q^2}}},\,p \ne q,{\text{ then }}\frac{{{a_6}}}{{{a_{21}}}}{\text{ equals}}$$
A.
$$\frac{{41}}{{11}}$$
B.
$$\frac{{7}}{{2}}$$
C.
$$\frac{{2}}{{7}}$$
D.
$$\frac{{11}}{{41}}$$
Answer :
$$\frac{{11}}{{41}}$$
Solution :
$$\eqalign{
& \frac{{\frac{p}{2}\left[ {2{a_1} + \left( {p - 1} \right)d} \right]}}{{\frac{q}{2}\left[ {2{a_1} + \left( {q - 1} \right)d} \right]}} = \frac{{{p^2}}}{{{q^2}}} \cr
& \Rightarrow \,\,\frac{{2{a_1} + \left( {p - 1} \right)d}}{{2{a_1} + \left( {q - 1} \right)d}} = \frac{p}{q} \cr
& \frac{{{a_1} + \left( {\frac{{p - 1}}{2}} \right)d}}{{{a_1} + \left( {\frac{{q - 1}}{2}} \right)d}} = \frac{p}{q} \cr
& {\text{For }}\frac{{{a_6}}}{{{a_{21}}}},\,\,p = 11,\,\,q = 41 \cr
& \Rightarrow \,\,\frac{{{a_6}}}{{{a_{21}}}} = \frac{{11}}{{41}} \cr} $$