Question

In a GP if the ${\left(2p\right)}^{th}$ term is $q^2$ and ${\left(2q\right)}^{th}$ term is $p^2$ where $p$ and $q\in N$ , find the$(p+q{)}^{th}$ term.$1$

Answer 1

Step $1$: Frame equations using given conditions:

Let $a$ be the first term of GP and $r$ be the common difference.

Given : ${\left(2p\right)}^{th}$ term is $q^2$ and ${\left(2q\right)}^{th}$ term is $p^2$.

Since, $n^{th}$ term of G.P. is given by $a{r}^{\left(n-1\right)}$

Thus,

$a{r}^{(2p-1)}=q^2$………………………………………… $\left(1\right)$

$a{r}^{(2q-1)}=p^2$………………………………………… $\left(2\right)$

Step $2$: Dividing equation $1$ by equation $2$:

$\begin{array}{rcl}\frac{a{r}^{(2p-1)}}{a{r}^{(2q-1)}}& =& \frac{q^2}{p^2}\\ & \Rightarrow & r^{(2p-1-2q+1)}=\frac{q^2}{p^2}\\ & \Rightarrow & r^{2(p-q)}={\left(\frac{q}{p}\right)}^2\\ & \Rightarrow & r={\left(\frac{q}{p}\right)}^{\frac{1}{(p-q)}}\end{array}$

Step $3$: Find the ${(p+q)}^{th}$ term of GP.

From equation $\left(1\right)$. $a=\frac{q^2}{r^{(2p-1)}}$

Since, $n^{th}$ term of G.P. is given by $a{r}^{\left(n-1\right)}$.

Thus, ${\left(p+q\right)}^{th}$term of GP :

$\begin{array}{rcl}{T}_{p+q}& =& a{r}^{(p+q-1)}\\ & \Rightarrow & T_{p+q}=\frac{q^2}{r^{(2p-1)}}{r}^{(p+q-1)}\\ & \Rightarrow & T_{p+q}=q^2{r}^{(p+q-1-2p+1)}\\ & \Rightarrow & T_{p+q}=q^2{r}^{-(p-q)}\\ & \Rightarrow & T_{p+q}=q^2{\left[{\left(\frac{q}{p}\right)}^{\frac{1}{(p-q)}}\right]}^{-(p-q)}\\ & \Rightarrow & T_{p+q}=q^2\left(\frac{p}{q}\right)\\ & \Rightarrow & T_{p+q}=pq\end{array}$ ; $\left[r={\left(\frac{q}{p}\right)}^{\frac{1}{(p-q)}}\right]$

Hence, the ${\left(p+q\right)}^{th}$term of GP is $pq$.