Let {a_1},{a_2}...,{a_{10}} be a G. | vedclass

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Question

$\frac{{{a_3}}}{{{a_1}}} = \frac{{{a_1}{r^2}}}{{{a_1}}} = {r^2}$

$ \Rightarrow {r^2} = 25$

Now $\frac{{{a_9}}}{{{a_5}}} = \frac{{{a_1}{r^8}

Vedclass · Accepted Answer

$5^4$

Vedclass · Answer

$\frac{{{a_3}}}{{{a_1}}} = \frac{{{a_1}{r^2}}}{{{a_1}}} = {r^2}$

$ \Rightarrow {r^2} = 25$

Now $\frac{{{a_9}}}{{{a_5}}} = \frac{{{a_1}{r^8}}}{{{a_1}{r^4}}} = {r^4} = {\left( {25} \right)^2} = {5^4}$

Let ${a_1},{a_2}...,{a_{10}}$ be a $G.P.$ If $\frac{{{a_3}}}{{{a_1}}} = 25,$ then $\frac {{{a_9}}}{{{a_{ 5}}}}$ equal

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