Question

Prove that the $(n+1{)}^{th}$ term of a G.P., of which the first term is $a$ and the third them $b$, is equal to the $(2n+1{)}^{th}$ term of a G.P. of which the first term is $a$ and the fifth term $b$.

Answer 1

Series 1
First term $=a$ and common ratio $=r$
Let $a_n$ be the $nth$ term of G.P., given by
$a_n=a{r}^{n-1}$

Third term = $a{r}^2=b$ ........ [Given]
$r=\sqrt{\frac{b}{a}}$
$(n+1{)}^{th}$ term = $a{r}^n=a{\left(\frac{b}{a}\right)}^{\frac{n}{2}}$ ......... $\left(1\right)$
Series 2
First term $=a$ and common ratio $=r^{\prime }$
Let $a_n^{\prime }$ be the $nth$ term of G.P., given by
$a_n^{\prime }=a{r^{\prime }}^{n-1}$
Fifth term $=a{r}^{\prime 4}=b$ ..... [Given]
$r^{\prime }={\left(\frac{b}{a}\right)}^{\frac{1}{4}}$
$(2n+1{)}^{th}$ term $=a{r^{\prime }}^{2n}=a{\left(\frac{b}{a}\right)}^{\frac{2n}{4}}$ ........ $\left(2\right)$
On comparing $\left(1\right)$ and $\left(2\right)$, we get
$(2n+1{)}^{th}$ term of series 2 = $(n+1{)}^{th}$ term of series 1
Hence proved.