Question -
Answer -
(i)┬а(2/3x тАУ 3/2x)20
We have,
(2/3x тАУ 3/2x)20┬аwhere,n = 20 (even number)
So the middle term is(n/2 + 1) = (20/2 + 1) = (10 + 1) = 11. ie., 11th┬аterm
Now,
T11┬а=T10+1
=┬а20C10┬а(2/3x)20-10┬а(3/2x)10
=┬а20C10┬а210/310┬а├Ч310/210┬аx10-10
=┬а20C10
Hence, the middle termis┬а20C10.
(ii)┬а(a/x + bx)12
We have,
(a/x + bx)12┬аwhere,n = 12 (even number)
So the middle term is(n/2 + 1) = (12/2 + 1) = (6 + 1) = 7. ie., 7th┬аterm
Now,
T7┬а= T6+1
= 924 a6b6
Hence, the middle termis 924 a6b6.
(iii)┬а(x2┬атАУ2/x)10
We have,
(x2┬атАУ2/x)10┬аwhere, n = 10 (even number)
So the middle term is(n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th┬аterm
Now,
T6┬а= T5+1
Hence, the middle termis -8064x5.
(iv)┬а(x/a тАУ a/x)10
We have,
(x/a тАУ a/x)┬а10┬аwhere,n = 10 (even number)
So the middle term is(n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th┬аterm
Now,
T6┬а= T5+1
Hence, the middle termis -252.