Question

Find the ${12}^{th}$ term of a $G.P$. whose $8^{th}$ term is $192$ and the common ratio is $2$.

Answer 1

Common ratio, $r=2$
Let $a$ be the first term of the G.P.
$\therefore a_8={ar}^{8-1}={ar}^7\Rightarrow {ar}^7=192\Rightarrow {a\left(2\right)}^7=192\Rightarrow {a\left(2\right)}^7={\left(2\right)}^6\left(3\right)\Rightarrow a=\frac{{\left(2\right)}^6\times 3}{{\left(2\right)}^7}=\frac{3}{2}\therefore a^{12}={ar}^{12-1}=\left(\frac{3}{2}\right){\left(2\right)}^{11}=\left(3\right){\left(2\right)}^{16}=3072$