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

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8. Sequences and Series

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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

A

$\frac{3}{4}$

B

$\frac{1}{2}$

C

$\frac{1}{3}$

D

$\frac{2}{5}$

Solution

(b) ${T_5} = a{r^4} = \frac{1}{3}$ …..$(i)$

and ${T_9} = a{r^8} = \frac{{16}}{{243}}$ …..$(ii)$

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}$.

Standard 11

Mathematics

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