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Q. If $a_{1}, a_{2}, a_{3}, \ldots, a_{10}$ is a geometric progression and $\frac{a_{3}}{a_{1}}=25$, then $\frac{a_{9}}{a_{5}}$ equals

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Solution:

$\frac{ a _{3}}{ a _{1}}=25$
$\frac{ ar ^{2}}{ a }=25$
$r ^{2}=5^{2}$
$\frac{ a _{4}}{ a _{5}}=\frac{ ar ^{8}}{ ar ^{4}}= r ^{4}=5^{4}$