Question:

Step 1: Find the two AMs between $a$ and $b$. The sequence $a, A_1, A_2, b$ is an AP with common difference $d = \frac{b-a}{3}$. $$A_1 = a + \frac{b-a}{3} = \frac{2a+b}{3}, \quad A_2 = a + \frac{2(b-a)}{3} = \frac{a+2b}{3}$$ Step 2: Find the two GMs between $a$ and $b$. The sequence $a, G_1, G_2, b$ is a GP with ratio $r = \left(\frac{b}{a}\right)^{1/3}$. $$G_1 = a \cdot \left(\frac{b}{a}\right)^{1/3} = a^{2/3}b^{1/3}, \quad G_2 = a \cdot \left(\frac{b}{a}\right)^{2/3} = a^{1/3}b^{2/3}$$ $$G_1 G_2 = a^{2/3}b^{1/3} \cdot a^{1/3}b^{2/3} = ab$$ Option (c) is correct. Step 3: Find the two HMs between $a$ and $b$. The sequence $a, H_1, H_2, b$ means $\frac{1}{a}, \frac{1}{H_1}, \frac{1}{H_2}, \frac{1}{b}$ is an AP. Common difference: $d' = \frac{\frac{1}{b} - \frac{1}{a}}{3} = \frac{a-b}{3ab}$. $$\frac{1}{H_1} = \frac{1}{a} + \frac{a-b}{3ab} = \frac{3b + a - b}{3ab} = \frac{2b+a}{3ab} \Rightarrow H_1 = \frac{3ab}{a+2b}$$ $$\frac{1}{H_2} = \frac{1}{a} + \frac{2(a-b)}{3ab} = \frac{3b + 2a - 2b}{3ab} = \frac{b+2a}{3ab} \Rightarrow H_2 = \frac{3ab}{2a+b}$$ Step 4: Check option (a): $A_1 H_2 = \frac{2a+b}{3} \cdot \frac{3ab}{2a+b} = ab$. Correct. Step 5: Check option (d): $A_2 H_1 = \frac{a+2b}{3} \cdot \frac{3ab}{a+2b} = ab$. Correct. Step 6: Option (b) states $A_1 H_2 = a^2 b^2$, but we found $A_1 H_2 = ab$. Incorrect.