Question

The nth term of a GP 18, -12, 8,...... is $\frac{512}{729}$. The value of n is .

• A. 7
• B. 8
• C. 9
• D. 10

Answer 1

The correct option is C 9

The given GP 18, -12, 8,... The first term of the GP is 18 and the common ratio is $r=\frac{-12}{18}=-\frac{2}{3}{a}_n=\frac{512}{729}\Rightarrow {\mathrm{ar}}^{n-1}=\frac{512}{729}\Rightarrow 18{\left(-\frac{2}{3}\right)}^{n-1}=\frac{512}{729}\Rightarrow {\left(-\frac{2}{3}\right)}^{n-1}=\frac{512}{729\times 18}=\frac{2^9}{3^6\times 2\times 3^2}\Rightarrow {\left(-\frac{2}{3}\right)}^{n-1}=\frac{2^8}{3^8}={\left(-\frac{2}{3}\right)}^8\Rightarrow n-1=8\Rightarrow n=9$