Let A₁ and A₂ be two arithmetic means and G₁, G₂, G₃ be three geometric
means of two distinct positive numbers. Then G₁⁴+G₂⁴+G₃⁴+G₁² G₃² is equal to :

Solution

Maths sir · 4 steps

Notice that G₂⁴ = G₁²G₃², and the whole expression folds into one perfect square.

1. Set up. Call the two numbers a and b. Then a, G₁, G₂, G₃, b is a GP and a, A₁, A₂, b is an AP. Terms equidistant from the ends of an AP have equal sums, so A₁ + A₂ = a + b.
2. Make a square. In a GP, any term squared equals the product of the terms equidistant from it, so G₂² = G₁G₃. Squaring gives G₂⁴ = G₁²G₃². The expression becomes G₁⁴ + 2G₁²G₃² + G₃⁴ = (G₁² + G₃²)².
3. Simplify the bracket. Since a, G₁, G₂ are consecutive terms, G₁² = aG₂. Since G₂, G₃, b are consecutive terms, G₃² = bG₂. Adding these, G₁² + G₃² = G₂(a + b) = G₂(A₁ + A₂).

   (G₁² + G₃²)² = G₂²(A₁ + A₂)² = (A₁ + A₂)² G₁G₃

4. So. The expression equals (A₁ + A₂)² G₁G₃, which is option (a).

The idea to keep. With several means between two numbers, pair terms from the two ends. In an AP the pair sums are equal, and in a GP the pair products are equal. Then look for a perfect square.

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