Question

Find the $n^{th}$ term of a G.P $\frac{3}{2},\frac{3}{4},\frac{3}{8},...$

Answer 1

The series $\frac{3}{2},\frac{3}{4},\frac{3}{8},...$ is in G.P

Here first term$a_1=a=\frac{3}{2}$

Common ratio$=r=\frac{a_2}{a_1}=\frac{\frac{3}{4}}{\frac{3}{2}}=\frac{3}{4}\times \frac{2}{3}=\frac{1}{2}$

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

$n^{th}$ term$=\frac{3}{2}\times {\left(\frac{1}{2}\right)}^{n-1}$

$n^{th}$ term$=\frac{3}{2}\times {\left(\frac{1}{2}\right)}^n{\left(\frac{1}{2}\right)}^{-1}$

$n^{th}$ term$=\frac{3}{2}\times {\left(\frac{1}{2}\right)}^n\times 2$

$n^{th}$ term$=3{\left(\frac{1}{2}\right)}^n$

$n^{th}$ term$=3\times (2{)}^{-n}$

Hence $n^{th}$ term is $3\times (2{)}^{-n}$