Find the middle term in the expansion of :

Solution:

(i) For $(3+x)^{6}$

$a=3, b=x$ and $n=6$

As $n$ is even, $\left(\frac{n+2}{2}\right)^{\text {th }}$ is the middle term

Therefore, the middle term $=\left(\frac{6+2}{2}\right)^{\text {th }}=\left(\frac{8}{2}\right)^{\text {th }}=(4)^{\text {th }}$

General term $\mathrm{t}_{\mathrm{r}+1}$ is given by,

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

Therefore, for $4^{\text {th }}, r=3$

Therefore, the middle term is

$t_{4}=t_{3+1}$

$=\left(\begin{array}{l}6 \\ 3\end{array}\right)(3)^{6-3}(\mathrm{x})^{3}$

$=\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \cdot(3)^{3}(\mathrm{x})^{3}$

$=(20) \cdot(27) \mathrm{x}^{3}$

$=540 \mathrm{x}^{3}$

(ii) For $\left(\frac{x}{3}+3 y\right)^{8}$,

$a=\frac{x}{3}, b=3 y$ and $n=8$

As $n$ is even, $\left(\frac{n+2}{2}\right)^{\text {th }}$ is the middle term

Therefore, the middle term $=\left(\frac{8+2}{2}\right)^{\text {th }}=\left(\frac{10}{2}\right)^{\text {th }}=(5)^{\text {th }}$

General term $\mathrm{t}_{\mathrm{r}+1}$ is given by,

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

Therefore, for $5^{\mathrm{th}}, \mathrm{r}=4$

Therefore, the middle term is

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

$=\left(\begin{array}{l}8 \\ 4\end{array}\right)\left(\frac{x}{3}\right)^{8-4}(3 y)^{4}$

$=\left(\begin{array}{l}8 \\ 4\end{array}\right)\left(\frac{x}{3}\right)^{4}(3)^{4}(y)^{4}$

$=\left(\begin{array}{l}8 \\ 4\end{array}\right) \frac{(x)^{4}}{(3)^{4}}(3)^{4}(y)^{4}$

$=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \cdot(x)^{4}(y)^{4}$

$=(70) \cdot x^{4} y^{4}$

(iii) For $\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$,

$a=\frac{x}{a} b=\frac{-a}{x}$ and $n=10$

As $n$ is even, $\left(\frac{n+2}{2}\right)^{\text {th }}$ is the middle term

Therefore, the middle term $=\left(\frac{10+2}{2}\right)^{\text {th }}=\left(\frac{12}{2}\right)^{\text {th }}=(6)^{\text {th }}$

General term $\mathrm{tr}+1$ is given by,

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

Therefore, for $6^{\text {th }}, \mathrm{r}=5$

Therefore, the middle term is

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

$=\left(\begin{array}{c}10 \\ 5\end{array}\right)\left(\frac{x}{a}\right)^{10-5}\left(\frac{-a}{x}\right)^{5}$

$=\left(\begin{array}{c}10 \\ 5\end{array}\right)\left(\frac{x}{a}\right)^{5}(-a)^{5}\left(\frac{1}{x}\right)^{5}$

$=\left(\begin{array}{c}10 \\ 5\end{array}\right) \frac{(x)^{5}}{(a)^{5}}(-a)^{5}\left(\frac{1}{x}\right)^{5}$

$=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \cdot(-1)$

$=-252$

(iv) For $\left(x^{2}-\frac{2}{x}\right)^{10}$

$a=x^{2}, \quad b=\frac{-2}{x}$ and $n=10$

As $n$ is even, $\left(\frac{n+2}{2}\right)^{\text {th }}$ is the middle term

Therefore, the middle term $=\left(\frac{10+2}{2}\right)^{\text {th }}=\left(\frac{12}{2}\right)^{\text {th }}=(6)^{\text {th }}$

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

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

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

Therefore, the middle term is

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

$=\left(\begin{array}{c}10 \\ 5\end{array}\right)\left(x^{2}\right)^{10-5}\left(\frac{-2}{x}\right)^{5}$

$=\left(\begin{array}{c}10 \\ 5\end{array}\right)\left(\mathrm{x}^{2}\right)^{5}(-2)^{5}\left(\frac{1}{\mathrm{x}}\right)^{5}$

$=\left(\begin{array}{c}10 \\ 5\end{array}\right) \frac{(\mathrm{x})^{10}}{(\mathrm{x})^{5}}(-32)$

$=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \cdot(-32)(\mathrm{x})^{5}$

$=-252(32) \mathrm{x}^{5}$

$=-8064 \mathrm{x}^{5}$