# Find the two middle terms in the expansion of :

Question:

Find the two middle terms in the expansion of :

$\left(\frac{p}{x}+\frac{x}{p}\right)^{9}$

Solution:

For $\left(\frac{\mathrm{p}}{\mathrm{x}}+\frac{\mathrm{x}}{\mathrm{p}}\right)^{9}$

$a=\frac{p}{x}, b=\frac{x}{p}$ and $n=9$

As n is odd, there are two middle terms i.e.

I. $\left(\frac{n+1}{2}\right)^{\text {th }}$ and II. $\left(\frac{n+3}{2}\right)^{\text {th }}$

General term $t_{r+1}$ is given by,

$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

I. The first middle term is $\left(\frac{\mathrm{n}+1}{2}\right)^{\text {th }}=\left(\frac{9+1}{2}\right)^{\text {th }}=\left(\frac{10}{2}\right)^{\text {th }}=(5)^{\text {th }}$

Therefore, for $5^{\text {th }}$ middle term, $r=4$

Therefore, the first middle term is

$\mathrm{t}_{5}=\mathrm{t}_{4+1}$

$=\left(\begin{array}{l}9 \\ 4\end{array}\right)\left(\frac{\mathrm{p}}{\mathrm{x}}\right)^{9-4}\left(\frac{\mathrm{x}}{\mathrm{p}}\right)^{4}$

$=\left(\begin{array}{l}9 \\ 4\end{array}\right)\left(\frac{\mathrm{p}}{\mathrm{x}}\right)^{5}(\mathrm{x})^{4}\left(\frac{1}{\mathrm{p}}\right)^{4}$

$=\left(\begin{array}{l}9 \\ 4\end{array}\right)\left(\frac{\mathrm{p}}{\mathrm{X}}\right)$

$=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \cdot(\mathrm{p}) \cdot(\mathrm{x})^{-1}$

$=126 \mathrm{p} \cdot \mathrm{x}^{-1}$+

II. The second middle term is $\left(\frac{\mathrm{n}+3}{2}\right)^{\text {th }}=\left(\frac{9+3}{2}\right)^{\text {th }}=\left(\frac{12}{2}\right)^{\text {th }}=(6)^{\text {th }}$

Therefore, for the $6^{\text {th }}$ middle term, $r=5$

Therefore, the second middle term is

$\mathrm{t}_{6}=\mathrm{t}_{5+1}$

$=\left(\begin{array}{l}9 \\ 5\end{array}\right)\left(\frac{\mathrm{p}}{\mathrm{X}}\right)^{9-5}\left(\frac{\mathrm{x}}{\mathrm{p}}\right)^{5}$

$=\left(\begin{array}{l}9 \\ 4\end{array}\right)\left(\frac{\mathrm{p}}{\mathrm{x}}\right)^{4}(\mathrm{x})^{5}\left(\frac{1}{\mathrm{p}}\right)^{5} \ldots \ldots \ldots\left(\because\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)\right]$

$=\left(\begin{array}{l}9 \\ 4\end{array}\right)\left(\frac{\mathrm{x}}{\mathrm{p}}\right)$

$=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \cdot\left(\frac{1}{\mathrm{p}}\right) \cdot(\mathrm{x})$

$=126\left(\frac{1}{\mathrm{p}}\right) \cdot(\mathrm{x})$