Question

If the $p^{th}$ term of a G. P. is $x$and $q^{th}$ term is $y$, then the $n^{th}$term is

• A. ${\left(\frac{x^{n-p}}{y^{n-q}}\right)}^{\frac{1}{\left(p-q\right)}}$
• B. ${\left(\frac{x^{n+q}}{y^{n+p}}\right)}^{\frac{1}{\left(p-q\right)}}$
• C. ${\left(\frac{x^{n-q}}{y^{n-p}}\right)}^{\frac{1}{\left(p-q\right)}}$
• D. ${\left(\frac{x^{n-q}}{y^{n-q}}\right)}^{\frac{1}{\left(p+q\right)}}$

Answer 1

The correct option is C

${\left(\frac{x^{n-q}}{y^{n-p}}\right)}^{\frac{1}{\left(p-q\right)}}$

Explanation of the correct option.

Step $1$: Find the value of $r$.

Given : $p^{th}$ term of a G. P. is $x$and $q^{th}$ term is $y$.

So, $p^{th}$ term $=a{r}^{p-1}$ [ $n^{th}$term of the G.P.$=a{r}^{n-1}$]

And $q^{th}$ term $=a{r}^{q-1}$

$\therefore x=a{r}^{p-1}$ and $y=a{r}^{q-1}$

By dividing $x$ by $y$, we get

$\frac{x}{y}=\frac{r^{p-1}}{r^{q-1}}$

$\Rightarrow \frac{x}{y}=r^{\left(p-1\right)-\left(q-1\right)}=r^{\left(p-q\right)}$

$\Rightarrow r={\left(\frac{x}{y}\right)}^{\frac{1}{\left\{p-q\right\}}}$

Step $2$: Find the required term.

Since, $n^{th}$the term of the G.P.is given by, $a{r}^{n-1}$

$t_n=a{r}^{n-1}$

It can be written as,

$$\begin{gathered} \Rightarrow t_n=a.r^{n-1}.\left(r^p\right).\left(r^{-p}\right) \\ \Rightarrow t_n=a.r^{p-1}.r^{n-p} \\ \Rightarrow t_n=x.{\left(\frac{x}{y}\right)}^{\frac{n-p}{p-q}} \\ \Rightarrow t_n=x.{\left(\frac{x^{n-p}}{y^{n-p}}\right)}^{\frac{1}{p-q}} \\ \Rightarrow t_n={\left(\frac{x^{p-q}.x^{n-p}}{y^{n-p}}\right)}^{\frac{1}{p-q}} \\ \Rightarrow t_n={\left(\frac{x^{n-q}}{y^{n-p}}\right)}^{\frac{1}{p-q}} \end{gathered}$$

Hence option $\left(C\right)$ is the correct option.