Find the middle term in the expansion of-i 23 x-32 - 20ii a x-b x 12iii - 2-2 - 10iv xa-ax 10

(i) Here, n = 20 nbsp; (Even number) nbsp; Therefore, the middle term is the n2+1th term, i.e., the 11th term. Now,T11=T10+1=C1020 #160; #160;23x20 10 ...

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Standard XI

Mathematics

Sequence

Find the midd...

Question

Find the middle term in the expansion of:
(i) ${(\frac{2}{3} x - \frac{3}{2 x})}^{20}$

(ii) ${(\frac{a}{x} + b x)}^{12}$

(iii) ${(x^{2} - \frac{2}{x})}^{10}$

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

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Solution

(i) Here,
n = 20 (Even number)
Therefore, the middle term is the $(\frac{n}{2} + 1)$ th term, i.e., the 11th term.
$Now, T_{11} = T_{10 + 1} = {}^{20}C_{10} {(\frac{2}{3} x)}^{20 - 10} {(\frac{3}{2 x})}^{10} = {}^{20}C_{10} \frac{2^{10}}{3^{10}} \times \frac{3^{10}}{2^{10}} x^{10 - 10} = {}^{20}C_{10}$

(ii) Here,
n = 12 (Even number)
Therefore, the middle term is the $(\frac{n}{2} + 1) th$ i.e. 7th term
$Now, T_{7} = T_{6 + 1} = {}^{12}C_{6} {(\frac{a}{x})}^{12 - 6} (b x)^{6} = {}^{12}C_{6} a^{6} b^{6} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2} a^{6} b^{6} = 924 a^{6} b^{6}$

(iii) Here,
n = 10 (Even number)
Therefore, the middle term is the $(\frac{n}{2} + 1) th$ i.e. 6th term
$Now, T_{6} = T_{5 + 1} = {}^{10}C_{5} (x^{2})^{10 - 5} {(\frac{- 2}{x})}^{5} = - \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2} \times 32 x^{5} = - 8064 x^{5}$

(iv) Here,
n = 10 (Even number)
Therefore, the middle term is the $(\frac{n}{2} + 1) th$ i.e. 6th term
$Now, T_{6} = T_{5 + 1} = {}^{10}C_{5} {(\frac{x}{a})}^{10 - 5} {(\frac{- a}{x})}^{5} = - \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2} = - 252$

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Find the middle term in the expansion of: (i) 23x-32x20 (ii) ax+bx12 (iii) x2-2x10 (iv) xa-ax10

Find the middle term in the expansion of:
(i) ${(\frac{2}{3} x - \frac{3}{2 x})}^{20}$

(ii) ${(\frac{a}{x} + b x)}^{12}$

(iii) ${(x^{2} - \frac{2}{x})}^{10}$

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