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# If the $\;10^{th}\;$ term of a $GP$ is $9$ and the $\;4^{th}\;$ term is $4$, then the $\;7^{th} \;$term is

$(a)\;\frac{9}{4}\qquad(b)\;7\qquad(c)\;6\qquad(d)\;\frac{4}{9}$

Explanation : $\;a_{4}=4=ar^3$
$a_{10}=9=ar^9$
Method 1 : $a_{7}=\sqrt{(a_{4}*a_{10})}\;(4,7,10\;are\;in\;AP)$
$a_{7}=\sqrt{9*4}=\sqrt{36}=6.$
Method 2 : $\frac{a_{10}}{a_{4}}=\frac{ar^9}{ar^3}=r^6\;=\frac{9}{4}$
$a_{4}=4=ar^3=\frac{3}{2}\;a$
$a=4*\;\frac{2}{3}=\frac{8}{3}$
Replacing value of a and $\;r^6\;in\;a_{7}=ar^6$
We get $\;\frac{8}{3}*\;(3/2)^2\;$
$a_{7}=\frac{8}{3}\;*\frac{3}{2}\;*\frac{3}{2}=2*3=6.$