# Find the middle term in the expansion of:

Question:

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

Solution:

(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,

$T_{6}=T_{5+1}$

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

$=-\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2}=-252$