Question

Find the middle terms in the expansion of:

(i) (x/3 + 9y)¹⁰

(ii) (3 – x³/6)⁷

(iii) (2ax – b/x²)¹²

(iv) (p/x + x/p)⁹

(v) (x/a – a/x)¹⁰

Answer 1

(i) (x/3 + 9y)¹⁰

Here we have,

(x/3 + 9y)¹⁰ where, n = 10 is an even number.

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

Now,

T₆ = T₅₊₁

Thus, the middle term is 61236x⁵y⁵.

(ii) (3 – x³/6)⁷

Here we have,

(3 – x³/6)⁷ where, n = 7 (odd number).

Therefore the middle terms are ((n + 1)/2) = ((7 + 1)/2) = 8/2 = 4 and

((n + 1)/2 + 1) = ((7 + 1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5

The terms are 4^th and 5^th.

Now,

T₄ = T₃₊₁

= ⁷C₃ (3)^7-3 (-x³/6)³

= -105/8 x⁹

And,

T₅ = T₄₊₁

= ⁹C₄ (3)^9-4 (-x³/6)⁴

Hence, the middle terms are -105/8 x⁹ and 35/48 x¹².

(iii) (2ax – b/x²)¹²

We have,

(2ax – b/x²)¹² where, n = 12 is an even number.

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

Now,

T₇ = T₆₊₁

Thus, the middle term is (59136a⁶b⁶)/x⁶.

(iv) (p/x + x/p)⁹

We have,

(p/x + x/p)⁹ where, n = 9 (odd number).

Therefore the middle terms are ((n + 1)/2) = ((9 + 1)/2) = 10/2 = 5 and

((n + 1)/2 + 1) = ((9 + 1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6

The terms are 5^th and 6^th.

Now,

T₅ = T₄₊₁

And,

T₆ = T₅₊₁

= ⁹C₅ (p/x)^9-5 (x/p)⁵

Thus, the middle terms are 126p/x and 126x/p.

(v) (x/a – a/x)¹⁰

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

Now,

T₆ = T₅₊₁

Thus, the middle term is -252.