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

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Question

(b) ${T_5} = a{r^4} = \frac{1}{3}$ &hellip;..$(i)$

and ${T_9} = a{r^8} = \frac{{16}}{{243}}$ &hellip;..$(ii)$

Solving $(i)$ and $(ii),$

we ge

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(b) ${T_5} = a{r^4} = \frac{1}{3}$ &hellip;..$(i)$

and ${T_9} = a{r^8} = \frac{{16}}{{243}}$ &hellip;..$(ii)$

Solving $(i)$ and $(ii),$

we get $r = \frac{2}{3}$ and $a = \frac{{27}}{{16}}$

Now ${4^{th}}$ term $ = a{r^3} = \frac{{{3^3}}}{{{2^4}}}.\frac{{{2^3}}}{{{3^3}}} = \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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