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If the $10^{\text {th }}$ term of a geometric progression is 9 and $4^{\text {th }}$ term is 4 , then its $7^{\text {th }}$ term is
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The correct answer is:
$6$
Accordingly, $a r^9=9$ and $a r^3=4$
$\Rightarrow r^3=\frac{3}{2}$ and $a=\frac{8}{3}$
$\therefore 7^{\text {th }}$ term i.e. $a r^6=\frac{8}{3}\left(\frac{3}{2}\right)^2=6$
Trick : $7^{\text {th }}$ term is equidistant from $10^{\text {th }}$ and $4^{\text {th }}$ so it will be $\sqrt{9 \times 4}=6$.
$\Rightarrow r^3=\frac{3}{2}$ and $a=\frac{8}{3}$
$\therefore 7^{\text {th }}$ term i.e. $a r^6=\frac{8}{3}\left(\frac{3}{2}\right)^2=6$
Trick : $7^{\text {th }}$ term is equidistant from $10^{\text {th }}$ and $4^{\text {th }}$ so it will be $\sqrt{9 \times 4}=6$.
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