If the 5^{th} term of a G.P. is frac{1}{3} an | vedclass

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(B) Let the first term be $a$ and the common ratio be $r$ for the $G.P.$The $n^{th}$ term of a $G.P.$ is given by $T_n = ar^{n-1}$.Given $T_5 = ar^4 =

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$\frac{1}{2}$

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(B) Let the first term be $a$ and the common ratio be $r$ for the $G.P.$The $n^{th}$ term of a $G.P.$ is given by $T_n = ar^{n-1}$.Given $T_5 = ar^4 = \frac{1}{3}$  --- $(i)$Given $T_9 = ar^8 = \frac{16}{243}$ --- $(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{ar^8}{ar^4} = \frac{16/243}{1/3}$$r^4 = \frac{16}{243} 	imes 3 = \frac{16}{81}$$r^4 = (\frac{2}{3})^4$,which implies $r = \frac{2}{3}$ (taking the positive root).Substituting $r = \frac{2}{3}$ into equation $(i)$:$a(\frac{2}{3})^4 = \frac{1}{3}$$a(\frac{16}{81}) = \frac{1}{3}$$a = \frac{1}{3} 	imes \frac{81}{16} = \frac{27}{16}$Now,the $4^{th}$ term $T_4 = ar^3$:$T_4 = \frac{27}{16} 	imes (\frac{2}{3})^3$$T_4 = \frac{27}{16} 	imes \frac{8}{27} = \frac{8}{16} = \frac{1}{2}$.

If the $5^{th}$ term of a $G.P.$ is $\frac{1}{3}$ and $9^{th}$ term is $\frac{16}{243}$,then the $4^{th}$ term will be

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