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 the third term from the beginning is :

Solution

Maths sir · 4 steps

The fifth term from the end is the fifth term of the reversed expansion, so the coefficients cancel.

1. Two fifth terms. Let a = 2^(1/4) and b = 3^(-1/4). From the beginning, T₅ = C(n, 4) aⁿ⁻⁴ b⁴. From the end, read the expansion of (b + a)ⁿ, so the fifth term is C(n, 4) bⁿ⁻⁴ a⁴.
2. Take the ratio. The C(n, 4) cancels, leaving aⁿ⁻⁸/bⁿ⁻⁸ = (a/b)ⁿ⁻⁸. Here a/b = 2^(1/4) × 3^(1/4) = 6^(1/4).

   6^((n-8)/4) = 6^(1/2) ⇒ (n-8)/4 = 1/2 ⇒ n = 10

3. Third term. T₃ = C(10, 2) a⁸ b² = 45 × 2² × 3^(-1/2) = 180/√3.
4. So. Rationalising, 180/√3 = 60√3. The answer is (b).

The idea to keep. The r-th term from the end of (a+b)ⁿ is the r-th term of (b+a)ⁿ. In the ratio of matching terms the coefficients cancel, leaving (a/b)ⁿ⁻²ʳ⁺².

Answer from the NTA final answer key. Solution by PrepWiser.