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Find the middle term in the expansion of : 

(i) (3 + x)6

(ii) \(\Big(\frac{x}{3}+3y\Big)^8\)

(iii) \(\Big(\frac{x}{a} - \frac{a}{x}\Big)^{10}\)

(iv)   \(\Big(x^2- \frac{2}{x}\Big)^{10}\)

1 Answer

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Best answer

(i) For (3 + x)6

a=3, b=x and n=6

As n is even, is the middle term 

Therefore, \(\Big(\frac{n+2}{2}\Big)^{th}\) the middle term 

Therefore, the middle term =   \(\Big(\frac{6+2}{2}\Big)^{th}\) =   \(\Big(\frac{8}{2}\Big)^{th}\) = (4)th

General term tr+1 is given by,

Therefore, for 4th , r=3 

tr+1 = (rn) an-rbr

Therefore, for 4th , r=3

Therefore, the middle term is

t4 = t3+1

a = \(\frac{x}{3}\) , b = 3y and n = 8

As n is even,   \(\Big(\frac{n+2}{2}\Big)^{th}\)  is the middle term 

Therefore, the middle term = \(\Big(\frac{8+2}{2}\Big)^{th}\) =   \(\Big(\frac{10}{2}\Big)^{th}\) = (5)th

General term tr+1 is given by,

tr+1 = (rn) an-rbr

Therefore, for 5th , r = 4 

Therefore, the middle term is

t5 = t4+1

(iii) For \(\Big(\frac{x}{a} - \frac{a}{x}\Big)^{10}\),

a = \(\frac{x}{a}\) , b =   \(\frac{-a}{x}\)  and n=10 

As n is even,     \(\Big(\frac{n+2}{2}\Big)^{th}\)  is the middle term

Therefore, the middle term = \(\Big(\frac{10+2}{2}\Big)^{th}\) =   \(\Big(\frac{12}{2}\Big)^{th}\) = (6)th

General term tr+1 is given by, 

tr+1 = (rn)an-rbr

Therefore, for 6th , r = 5 

Therefore, the middle term is

t6 = t5+1

= -252 

(iv) For , \(\Big(x^2- \frac{2}{x}\Big)^{10}\)

a = x2 , b = \(\frac{-2}{x}\) and n = 10 

As n is even,  \(\Big(\frac{n+2}{2}\Big)^{th}\)  is the middle term

Therefore, the middle term =  \(\Big(\frac{10+2}{2}\Big)^{th}\) =   \(\Big(\frac{12}{2}\Big)^{th}\) = (6)th

General term tr+1 is given by,

tr+1 = (rn) an-rbr

Therefore, for the 6th middle term, r = 5 

Therefore, the middle term is

t6 = t5+1

= -252 (32) x5 

= -8064 x5

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