Question:

<p><strong>Step 1:</strong> Let the two positive numbers be $a$ and $b$ where $a > b$.</p><p>Given: AM of two numbers = 2</p><p>$$\frac{a+b}{2} = 2 \implies a + b = 4$$</p><p><strong>Step 2:</strong> When the larger number is increased by 1, the new numbers are $(a+1)$ and $b$.</p><p>The GM of these new numbers equals the original AM (which is 2):</p><p>$$\sqrt{(a+1) \cdot b} = 2$$</p><p>Squaring both sides:</p><p>$$(a+1)b = 4$$</p><p><strong>Step 3:</strong> Expand equation from Step 2:</p><p>$$ab + b = 4$$</p><p><strong>Step 4:</strong> From Step 1: $a = 4 - b$</p><p>Substitute into the equation from Step 3:</p><p>$$(4-b)b + b = 4$$</p><p>$$4b - b^2 + b = 4$$</p><p>$$5b - b^2 = 4$$</p><p>$$b^2 - 5b + 4 = 0$$</p><p>$$(b-1)(b-4) = 0$$</p><p>So $b = 1$ or $b = 4$</p><p><strong>Step 5:</strong> Since $a > b$ and $a + b = 4$:</p><p>If $b = 1$, then $a = 3$ ✓ (and $a > b$)</p><p>If $b = 4$, then $a = 0$ ✗ (not positive)</p><p>Therefore: $a = 3, b = 1$</p><p><strong>Step 6:</strong> Verify: GM of $(3+1)$ and $1$ is $\sqrt{4 \cdot 1} = 2$ ✓ (equals original AM)</p><p><strong>Step 7:</strong> Calculate HM of the original numbers $a=3$ and $b=1$:</p><p>$$HM = \frac{2ab}{a+b} = \frac{2 \cdot 3 \cdot 1}{3+1} = \frac{6}{4} = \frac{3}{2}$$</p><p><strong>∴ Answer: a</strong></p>