Find the two middle terms in the expansion of :

For $\left(x^{2}+a^{2}\right)^{5}$

$a=x^{2}, b=a^{2}$ and $n=5$

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

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

I. The first, middle term is $\left(\frac{n+1}{2}\right)^{\text {th }}=\left(\frac{5+1}{2}\right)^{\text {th }}=\left(\frac{6}{2}\right)^{\text {th }}=(3)^{\text {rd }}$

Therefore, for the $3^{\text {rd }}$ middle term, $r=2$

Therefore, the first middle term is

$t_{3}=t_{2+1}$

$=\left(\begin{array}{l}5 \\ 2\end{array}\right)\left(\mathrm{x}^{2}\right)^{5-2}\left(\mathrm{a}^{2}\right)^{2}$

$=\left(\begin{array}{l}5 \\ 2\end{array}\right)\left(\mathrm{x}^{2}\right)^{3}(\mathrm{a})^{4}$

$=\left(\begin{array}{l}5 \\ 2\end{array}\right)(\mathrm{x})^{6}(\mathrm{a})^{4}$

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

$=10 . \mathrm{a}^{4} \cdot \mathrm{x}^{6}$

II. The second middle term is $\left(\frac{\mathrm{n}+3}{2}\right)^{\text {th }}=\left(\frac{5+3}{2}\right)^{\text {th }}=\left(\frac{8}{2}\right)^{\text {th }}=(4)^{\text {th }}$

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

Therefore, the second middle term is

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

$=\left(\begin{array}{l}5 \\ 3\end{array}\right)\left(\mathrm{x}^{2}\right)^{5-3}\left(\mathrm{a}^{2}\right)^{3}$

$=\left(\begin{array}{l}5 \\ 3\end{array}\right)\left(\mathrm{x}^{2}\right)^{2}(\mathrm{a})^{6}$

$=\left(\begin{array}{l}5 \\ 2\end{array}\right)(\mathrm{x})^{4}(\mathrm{a})^{6} \ldots \ldots \ldots\left(\because\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)\right]$

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

$=10 . \mathrm{a}^{6} \cdot \mathrm{x}^{4}$