For x ≥ 0, the least value of K, for which 4¹⁺ˣ+4¹⁻ˣ, K/2, 16ˣ+16⁻ˣ are three consecutive terms of an A.P., is equal to :

Solution

Maths sir · 4 steps

Write the A.P. condition first, then see that everything is a function of 4ˣ + 4⁻ˣ.

1. A.P. condition. For three consecutive terms of an A.P., twice the middle term equals the sum of the outer two. Here 2 × K/2 = K, so K = 4¹⁺ˣ+4¹⁻ˣ+16ˣ+16⁻ˣ.
2. One substitution. Put t = 4ˣ + 4⁻ˣ. Then 4¹⁺ˣ+4¹⁻ˣ = 4t and 16ˣ+16⁻ˣ = (4ˣ+4⁻ˣ)² - 2 = t² - 2.

   K = t² + 4t - 2 = (t+2)² - 6

3. Range of t. By AM–GM, 4ˣ + 4⁻ˣ ≥ 2, with equality at x = 0, which is allowed since x ≥ 0. So t ≥ 2, and on this range K = (t+2)² - 6 increases with t.
4. So. The least value is at t = 2: K = 16 - 6 = 10. The answer is (a).

The idea to keep. When terms like aˣ + a⁻ˣ and a²ˣ + a⁻²ˣ appear together, set t = aˣ + a⁻ˣ ≥ 2. The squared one becomes t² - 2, and the problem turns into a quadratic in t.

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