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# 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 x^{2} and x^{3} 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!) 3^{7} a^{2} = 9!/ (3! 6!) 3^{6} a^{3}

9! / (2! 7!) × (3! 6!) / 9! = (3^{6} a^{3}) / (3^{7} a^{2})

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

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