Question

If geometric mean and harmonic mean of two numbers are \(16\) and \(\frac{64}{5}\) respectively, then \(a:b\) is:

• A. \(4:1\)
• B. \(3:2\)
• C. \(2:3\)
• D. \(1:4\)

Hint:

Use \(GM^2 = ab\) and \(HM = \frac{2ab}{a+b}\) to solve for \(a\) and \(b\).

Answer 1

The Correct Option is A

Concept: For two numbers \(a\) and \(b\): \[ GM = \sqrt{ab} = 16 \quad \Rightarrow \quad ab = 256 \] \[ HM = \frac{2ab}{a+b} = \frac{64}{5} \]

Step 1: Find \(a+b\). \[ \frac{2 \times 256}{a+b} = \frac{64}{5} \quad \Rightarrow \quad \frac{512}{a+b} = \frac{64}{5} \] \[ 512 \times 5 = 64(a+b) \quad \Rightarrow \quad 2560 = 64(a+b) \quad \Rightarrow \quad a+b = 40 \]

Step 2: Solve for \(a\) and \(b\). We have: \[ a+b = 40,\quad ab = 256 \] The quadratic \(t^2 - 40t + 256 = 0\) gives: \[ t = \frac{40 \pm \sqrt{1600 - 1024}}{2} = \frac{40 \pm \sqrt{576}}{2} = \frac{40 \pm 24}{2} \] \[ t = 32 \quad \text{or} \quad t = 8 \] Thus, the numbers are 32 and 8. Ratio \(a:b = 32:8 = 4:1\).