(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