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If the $5^{\text {th }}$ term of a G.P. is $\frac{1}{3}$ and $9^{\text {th }}$ term is $\frac{16}{243}$, then the $4^{\text {th }}$ term will be
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The correct answer is:
$\frac{1}{2}$
$T_5=a r^4=\frac{1}{3} \ldots(i)$
and $T_9=a r^8=\frac{16}{243} \ldots(ii)$
Solving (i) and (ii), we get$r=\frac{2}{3}$ and $a=\frac{27}{16}$
Now $4^{\text {th }}$ term $=a r^3=\frac{3^3}{2^4} \cdot \frac{2^3}{3^3}=\frac{1}{2}$
and $T_9=a r^8=\frac{16}{243} \ldots(ii)$
Solving (i) and (ii), we get$r=\frac{2}{3}$ and $a=\frac{27}{16}$
Now $4^{\text {th }}$ term $=a r^3=\frac{3^3}{2^4} \cdot \frac{2^3}{3^3}=\frac{1}{2}$
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