Find the middle term (terms) in the expansion of

a. Given $\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$

Here index $n=10$ which is even number.

So, there is one middle term which is $(10 / 2+1)^{\text {th }}$ term that is $6^{\text {th }}$ term

$\therefore \quad T_{6}=T_{5+1}={ }^{10} C_{5}\left(\frac{x}{a}\right)^{10-5}\left(\frac{-a}{x}\right)^{5}$

$=-{ }^{10} C_{5}\left(\frac{x}{a}\right)^{5}\left(\frac{a}{x}\right)^{5}$

$=-\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 \times 4 \times 3 \times 2 \times 1}\left(\frac{x}{a}\right)^{5}\left(\frac{x}{a}\right)^{-5}=-252$

b. Given $\left(3 x-\frac{x^{3}}{6}\right)^{9}$

Here index $n=9$ which is odd

So, there is one middle term which is $(9 / 2+1)^{\text {th }}$ term that is $5^{\text {th }}$ term and $6^{\text {th }}$ term

$\therefore \quad T_{5}=T_{4+1}={ }^{9} C_{4}(3 x)^{9-4}\left(-\frac{x^{3}}{6}\right)^{4}$

$=\frac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !} 3^{5} x^{5} x^{12} 6^{-4}$

$=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \frac{3^{5}}{3^{4} \times 2^{4}} x^{17}=\frac{189}{8} x^{17}$

And $6^{\text {th }}$ term,

$T_{6}=T_{5+1}={ }^{9} C_{5}(3 x)^{9-5}\left(-\frac{x^{3}}{6}\right)^{5}$

$=-\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1} \cdot 3^{4} \cdot x^{4} \cdot x^{15} \cdot 6^{-5}=-\frac{-21}{16} x^{19}$