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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 · 2n/4 / 2 · (1/3)
= ⁿC₄ · 2n/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 · 2n/4 · 1/6] / [ⁿC₄ · 6 · 3-n/4] = √6 / 1
2n/4 · 3n/4 = 36 · √6
(2 · 3)n/4 = 62 · 61/2
6n/4 = 65/2
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
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