Question

In a GP., if $(2p{)}^{th}$ term is $q^2$ and $(2q{)}^{th}$ term is $p^2$ where $p$ and $q$ $\in N$, then its $(p+q{)}^{th}$ term is

• A. $pq$
• B. $p^2{q}^2$
• C. $\frac{1}{2}{p}^2{q}^2$
• D. $\frac{1}{4}{p}^3{q}^3$

Answer 1

The correct option is A $pq$

$T_{2p}=a{r}^{(2p-1)}=q^2$

$T_{2q}=a{r}^{2q-1)}=p^2$

$\frac{a{r}^{2p-1}}{a{r}^{2q-1}}=\frac{q^2}{p^2}$

$\Rightarrow r^{2p-2q}=q^2/p^2$

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

Now, $a{r}^{(2p-1)}=q^2$

$\Rightarrow a=\frac{q^2}{r^{2p-1}}$

Thus, $T_{p+q}=a{r}^{p+q-1}=\frac{q^2{r}^{p+q-1}}{r^{2p-1}}$

$=q^2{r}^{q-p}$

$=q^2{\left(\frac{q}{p}\right)}^{\frac{1}{p-q}\cdot q-p}$

$=q^2{\left(\frac{q}{p}\right)}^{-1}=q^2\frac{p}{q}$

$=pq$.