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Question

Find the middle term in the expansion of:

(i)(23x32x)20

(ii)(ax+bx)12

(iii)(x22x)10

(iv)(xaax)10

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Solution

(i)(23x32x)20

Here, n=20 which is an even number so, (202+1)thi.e.,11th term is the middle term We know that,

Tn=Tr+1=(1)rnCrxnryrn=20,r=10,x=23x,Y=23x

T11=T10+1=(1)1020C10(23x)10(32x)10

=20C10210310×310210×x10x10=20C10

(ii)(ax+bx)12

Here, n =12, which even number

So,(122+1)th term i.e., 7th term is the middle term.

Hence, the middle term =T7=T6+1

T7=T6+1=12C6×(ax)126×(bx)6

=12C6(ax)6×(bx)6

=12!(126)!6!×a6x6×b6×6

=12×11×10×9×8×7×6!(6×5×4×3×2×1)×a6b6

=924×a6b6

The middle term =924×a6b6.

(iii)(x22x)10Here,n=10

(n2+1)th=(102+1)th=6th term is the middle term.

The term formula is

Tn=Tr+1=(1)rnC5xnryr

T6=T5+1=(1)510C5(x2)105(2x)5

=10C5x201025x5

=10×9×8×7×65×4×3×2×25x5=8064x5

(iv)(xaax)10

Here n =10, which is even, therefore it has 11 terms

middletermis(n2+1)=6A term

Tn=Tr+1=(1)rnCrxnryr

T6=T5+1=(1)510C5(xa)105(xa)5

=10!5!5!×x5a3×a5×x5

=-252


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