Browse Questions

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

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

Answer : (a) $\frac{13}{27}$
Explanation : $\; S_{p}=\frac{p}{2}\;[2a+(p-1)d]$
$S_{q}=\frac{q}{2}\;[2a+(q-1)d]$
$\frac{S_{p}}{S_{q}}=\frac{a+(\frac{p-1}{2})d}{a+(\frac{q-1}{2})d}=\frac{p}{q}$
$put \;p=13\quad\;q=27$
$\frac{p}{q}=\frac{13}{27}.$