Let $a_{1}+a_{2}+a_{3}+\;......$ be terms of an AP. If $\;\frac{S_{p}}{S_{q}}=\frac{p^2}{q^2}\;then\;\frac{a_{7}}{a_{14}} =$

Question

$(a)\;\frac{13}{27}\qquad(b)\;\frac{27}{13}\qquad(c)\;\frac{6}{13}\qquad(d)\;\frac{13}{6}$