Question

which term of the G.P.
A) $2,8,32,....is512?$

B) $\sqrt{3},3,3\sqrt{3}......is729?$

C) $\frac{1}{3},\frac{1}{9},\frac{1}{27}.......is\frac{1}{2187}?$

Answer 1

$1$

$2,8,32\dots ..512$

$a=2r=4$

$a_n={ar}^{(n-1)}$

$\left(514\right)=2{\left(4\right)}^{(n-1)}$

$256={\left(4\right)}^{(n-1)}$

${\left(4\right)}^4=4^{(n-1)}\to n-1=4\to n=5$

$\therefore 512$ is the fifth term of the $G.P$

$2$

$\sqrt{3},3,3\sqrt{3},9\dots 729$

$a=\sqrt{3}r=\sqrt{3}$

$a_n={ar}^{(n-1)}$

$729=\sqrt{3}\times {\left(\sqrt{3}\right)}^{n-1}$

$3^6={\left(\sqrt{3}\right)}^n\to 3^6=3^{n/2}\to n=12$

$\therefore 729$ is the $12th$ term of the G.P.

$3$

$\frac{1}{3},\frac{1}{9},\frac{1}{27}\dots .\frac{1}{2187}$

$a=\frac{1}{3},r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{3}$

$a_n={ar}^{(n-1)}\to \frac{1}{2187}=\left(\frac{1}{3}\right){\left(\frac{1}{3}\right)}^{n-1}$

$\to 2178=3^n\to 3^7=3^n$

$\to =7$

$\therefore \frac{1}{2187}$ is the $7th$ term of the G.P.