Question

Which term of the progression 18, - 12, 8, .... is $\frac{512}{729}$ ?

Answer 1

18, - 12, 8, ..... is $\frac{512}{729}$

a = 18, n = ?, $t_n=\frac{512}{729}$.

$r=\frac{t_{n-1}}{t_n}=\frac{t_2}{t_1}=\frac{-12}{18}=\frac{-2}{3}$

Also,

$t_n=a{r}^{n-1}$

$\Rightarrow \frac{512}{729}=\left(18\right){\left(\frac{-2}{3}\right)}^{n-1}$

$\Rightarrow \frac{2^9}{36}\times \frac{1}{2\times 3^2}={\left(\frac{-2}{3}\right)}^{n-1}$

$\Rightarrow {\left(\frac{2}{3}\right)}^8=(-1{)}^{n-1}={\left(\frac{2}{3}\right)}^{n-1}$

$\Rightarrow n=9$

Thus, the 9th term of the given G.P. is $\frac{512}{729}$.