Question

Find the middle term in the expansion of:

(i) (2/3x – 3/2x)²⁰

(ii) (a/x + bx)¹²

(iii) (x² – 2/x)¹⁰

(iv) (x/a – a/x)¹⁰

Answer 1

(i) (2/3x – 3/2x)²⁰

Now, we have,

(2/3x – 3/2x)²⁰ where, n = 20 (even number)

Therefore the middle term is (n/2 + 1) = (20/2 + 1) = (10 + 1) = 11. ie., 11^th term

Then,

T₁₁ = T₁₀₊₁

= ²⁰C₁₀ (2/3x)^20-10 (3/2x)¹⁰

= ²⁰C₁₀ 2¹⁰/3¹⁰ × 3¹⁰/2¹⁰ x^10-10

= ²⁰C₁₀

Thus, the middle term is ²⁰C₁₀.

(ii) (a/x + bx)¹²

Now, we have,

(a/x + bx)¹² where, n = 12 (even number)

Therefore the middle term is (n/2 + 1) = (12/2 + 1) = (6 + 1) = 7. ie., 7^th term

Then,

T₇ = T₆₊₁

= 924 a⁶b⁶

Thus, the middle term is 924 a⁶b⁶.

(iii) (x² – 2/x)¹⁰

Now, we have,

(x² – 2/x)¹⁰ where, n = 10 (even number)

Therefore the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6^th term

Then,

T₆ = T₅₊₁

Thus, the middle term is -8064x⁵.

(iv) (x/a – a/x)¹⁰

Now we have,

(x/a – a/x) ¹⁰ where, n = 10 (even number)

Therefore the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6^th term

Then,

T₆ = T₅₊₁

Thus, the middle term is -252.