Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (∜2 + 1/∜3)ⁿ is √6 : 1

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.

Using this, in the expansion of (∜2 + 1/∜3)ⁿ,

• The fifth term from the beginning is,
  T₅ =  ⁿC₄ (∜2)^{n - 4} (1/∜3)⁴
  = ⁿC₄ · ((∜2)ⁿ / (∜2)⁴ · (1/3)
  = ⁿC₄ · 2^n/4 / 2 · (1/3)
  = ⁿC₄ · 2^n/4 · 1/6
   
• The fifth term from the end is,
  ⁿCₙ₋₄ (∜2)⁴ (1/∜3)^{n - 4}
  We know that ⁿCₙ₋₄ = ⁿC₄
  = ⁿC₄ · (2) · 3^-n/4 · (∜3)⁴
  = ⁿC₄ · 6 · 3^-n/4

It is given that the ratio of these two terms is √6 : 1.

[ⁿC₄ · 2^n/4 · 1/6] / [ⁿC₄ · 6 · 3^-n/4] = √6 / 1

2^n/4 · 3^n/4 = 36 · √6

(2 · 3)^n/4 = 6² · 6^1/2

6^n/4 = 6^5/2

n/4 = 5/2

n = 5/2 × 4 = 10

NCERT Solutions Class 11 Maths Chapter 8 Exercise ME Question 8

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (∜2 + 1/∜3)ⁿ is √6 : 1.

Summary:

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (∜2 + 1/∜3)ⁿ is √6 : 1, then n = 10