Find a if the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are equal

Solution:

It is known that (r + 1)^th term, (Tᵣ ₊ ₁) in the binomial expansion of (a + b)ⁿ is given by Tᵣ ₊ ₁ = ⁿCᵣ aⁿ ⁻ ʳ bʳ.

In the expansion of (3 + ax)⁹,

Tᵣ ₊ ₁ = ⁹Cᵣ (3)⁹ ⁻ ʳ (ax)ʳ

= ⁹Cᵣ (3)⁹ ⁻ ʳ aʳ xʳ

Using this, the coefficients of x² and x³ are ⁹C₂ (3)⁹ ⁻ ² a² and ⁹C₃ (3)⁹ ⁻ ³ a³ respectively. 

It is given that these coefficients are equal. So

⁹C₂ (3)⁹ ⁻ ² a² = ⁹C₃ (3)⁹ ⁻ ³ a³

Using nCr formula,

9! / (2! 7!) 3⁷ a² = 9!/ (3! 6!) 3⁶ a³

9! / (2! 7!) × (3! 6!) / 9! = (3⁶ a³) / (3⁷ a²)

3/7 = a/3

a = 9/7

NCERT Solutions Class 11 Maths Chapter 8 Exercise ME Question 2

Find a if the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are equal

Summary:

The value of a if the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are equal is 9/7