Question -
Answer -
(i) (2/3x – 3/2x)20
We have,
(2/3x – 3/2x)20 where,n = 20 (even number)
So the middle term is(n/2 + 1) = (20/2 + 1) = (10 + 1) = 11. ie., 11th term
Now,
T11 =T10+1
= 20C10 (2/3x)20-10 (3/2x)10
= 20C10 210/310 ×310/210 x10-10
= 20C10
Hence, the middle termis 20C10.
(ii) (a/x + bx)12
We have,
(a/x + bx)12 where,n = 12 (even number)
So the middle term is(n/2 + 1) = (12/2 + 1) = (6 + 1) = 7. ie., 7th term
Now,
T7 = T6+1
= 924 a6b6
Hence, the middle termis 924 a6b6.
(iii) (x2 –2/x)10
We have,
(x2 –2/x)10 where, n = 10 (even number)
So the middle term is(n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle termis -8064x5.
(iv) (x/a – a/x)10
We have,
(x/a – a/x) 10 where,n = 10 (even number)
So the middle term is(n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle termis -252.