Find the middle term in the expansion of:
(i) $\left(\frac{2}{3} x-\frac{3}{2 x}\right)^{20}$
(ii) $\left(\frac{a}{x}+b x\right)^{12}$
(iii) $\left(x^{2}-\frac{2}{x}\right)^{10}$
(iv) $\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$
(i) Here,
n = 20 (Even number)
Therefore, the middle term is the $\left(\frac{n}{2}+1\right)$ th term, i.e., the 11 th term.
Now,
$T_{11}=T_{10+1}$
$={ }^{20} C_{10}\left(\frac{2}{3} x\right)^{20-10}\left(\frac{3}{2 x}\right)^{10}$
$={ }^{20} C_{10} \frac{2^{10}}{3^{10}} \times \frac{3^{10}}{2^{10}} x^{10-10}$
$={ }^{20} C_{10}$
(ii) Here,
n = 12 (Even number)
Therefore, the middle term is the $\left(\frac{n}{2}+1\right)$ th i.e. 7 th term
Now,
$T_{7}=T_{6+1}$
$={ }^{12} C_{6}\left(\frac{a}{x}\right)^{12-6}(b x)^{6}$
$={ }^{12} C_{6} a^{6} b^{6}$
$=\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2} a^{6} b^{6}$
$=924 a^{6} b^{6}$
(iii) Here,
n = 10 (Even number)
Therefore, the middle term is the $\left(\frac{n}{2}+1\right)$ th i.e. 6th term
Now,
$T_{6}=T_{5+1}$
$={ }^{10} C_{5}\left(x^{2}\right)^{10-5}\left(\frac{-2}{x}\right)^{5}$
$=-\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2} \times 32 x^{5}$
$=-8064 x^{5}$
(iv) Here,
n = 10 (Even number)
Therefore, the middle term is the $\left(\frac{n}{2}+1\right)$ th i.e. 6th term
Now,
$T_{6}=T_{5+1}$
$={ }^{10} C_{5}\left(\frac{x}{a}\right)^{10-5}\left(\frac{-a}{x}\right)^{5}$
$=-\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2}=-252$