Question

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

• A. $9^{th}$
• B. ${10}^{th}$
• C. $8^{th}$
• D. $Noneofthese$

Answer 1

The correct option is A $9^{th}$

Let us consider the problem:

$a=18,r=\frac{-12}{18}=\frac{-2}{3}$

$T_n=a{r}^{n-1}\frac{512}{729}=18\times {\left(\frac{-2}{3}\right)}^{(n-1)}$

${\left(\frac{-2}{3}\right)}^{(n-1)}=\frac{512}{18\times 729}=\frac{2^9}{2\times 3^2\times 9^3}=\frac{2^8}{3^2\times 3^6}$

$=\frac{2^8}{3^8}$

$n-1=8n=9$.