Question

The $6^{th}$ and ${13}^{th}$ term of a G.P are respectively equal to $24$ and $\frac{3}{16}$.Find the ${25}^{th}$ term.

Answer 1

$n^{th}$ term of a G.P$t_n=a{r}^{n-1}$

$6^{th}$ term of a G.P$t_6=a{r}^{6-1}=a{r}^5=24\dots \left(1\right)$

${13}^{th}$ term of a G.P$t_{13}=a{r}^{13-1}=a{r}^{12}=\frac{3}{16}\dots \left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$

$\Rightarrow \frac{a{r}^5}{a{r}^{12}}=\frac{24}{\frac{3}{16}}=\frac{24\times 16}{3}=8\times 16$

$\Rightarrow \frac{1}{r^7}=2^3\times 2^4=2^7$

$\Rightarrow r^7=\frac{1}{2^7}$

$\therefore r=\frac{1}{2}$

We have $a{r}^5=24$

$\Rightarrow a=\frac{24}{r^5}$

$\Rightarrow a=\frac{24}{{\left(\frac{1}{2}\right)}^5}=24\times 32=768$

$\therefore a=768,r=\frac{1}{2}$

We know that, $t_n=a{r}^{(n-1)}$

$\therefore {25}^{th}$ term of a G.P$t_{25}$$=768\times {\left(\frac{1}{2}\right)}^{25-1}$

$t_{25}=\frac{768}{2^{24}}$

$t_{25}=\frac{3}{2^{16}}$

$\therefore {25}^{th}$ term is $3\times 2^{-16}$