# Find the middle terms in the expansion of (x/3 + 9y)¹⁰

**Solution:**

It is known that in the binomial expansion of (a + b)ⁿ, when n is even in the middle term is (n/2 + 1)ᵗʰ term.

Therefore the middle term in the expansion of (x/3 + 9y)¹⁰ is (10/2 + 1) = 6ᵗʰ term.

T₆ = T₅ ₊ ₁

= ¹⁰C₅ (x/3)¹⁰ ⁻ ⁵ (9y)⁵

= [10!/(5!)(5!)] × (x⁵/3⁵) × (9⁵) x (y⁵)

= [(10 × 9 × 8 × 7 × 6 × 5!) /(5 × 4 × 3 × 2 × 1 × 5!)] × (1/3⁵) × (3¹⁰) × (x⁵)(y⁵) (as 9⁵ = (3^{2})^{5} = 3^{10})

= 252 × 3⁵ × x⁵ × y⁵

= 61236 x⁵ y⁵

NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.2 Question 8

## Find the middle terms in the expansion of (x/3 + 9y)¹⁰.

**Summary:**

Using the binomial theorem, middle terms in the expansion of (x/3 + 9y)¹⁰ are to be found. We have found that it is 61236x⁵ y⁵

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