Question

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

• A. $7^{th}$
• B. $9^{th}$
• C. ${11}^{th}$
• D. ${13}^{th}$

Answer 1

Answer: (B)

$9^{th}$

$18,-12,8$

$=\frac{512}{729}=term$

$N^{th}$ term of $GP=a{r}^{n-1}$

$\Rightarrow \frac{512}{729}=18{\left(\frac{-12}{18}\right)}^{n-1}$

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

$\Rightarrow \frac{2^8}{2^8}=\frac{256}{9\times 9\times 9\times 9}={\left(\frac{-2}{3}\right)}^{n-1}$

$\Rightarrow n-1=8$

$\Rightarrow n=9$