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Question

Find the middle term in the expansion of:
(i) 23x-32x20

(ii) ax+bx12

(iii) x2-2x10

(iv) xa-ax10

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Solution

(i) Here,
n = 20 (Even number)
Therefore, the middle term is the n2+1th term, i.e., the 11th term.
Now,T11=T10+1=C1020 23x20-10 32x10=C1020 210310×310210x10-10=C1020

(ii) Here,
n = 12 (Even number)
Therefore, the middle term is the n2+1th i.e. 7th term
Now,T7=T6+1=C612 ax12-6 (bx)6=C612 a6 b6 =12×11×10×9×8×76×5×4×3×2a6 b6=924 a6b6

(iii) Here,
n = 10 (Even number)
Therefore, the middle term is the n2+1th i.e. 6th term
Now,T6=T5+1=C510 (x2)10-5 -2x5=-10×9×8×7×65×4×3×2×32x5=-8064 x5

(iv) Here,
n = 10 (Even number)
Therefore, the middle term is the n2+1th i.e. 6th term
Now,T6=T5+1 =C510 xa10-5 -ax5 =-10×9×8×7×65×4×3×2=-252

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