# 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} = 6^{2} · 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