Let one AM $a$ and two GMs $g_1$ and $g_2$ be inserted between $b$ and $c$. Then $\dfrac{g_1^3 + g_2^3}{abc} =$
Step-by-Step Solution
Key Concept: Use the definitions: $b+c=2a$ (AM), and $g_1 = b(c/b)^{1/3}$, $g_2 = b(c/b)^{2/3}$ (GMs). Compute $g_1^3=b^2c$ and $g_2^3=bc^2$.
$g_1^3=b^2c$, $g_2^3=bc^2$. Sum $= bc(b+c)=2abc$. Hence $\frac{g_1^3+g_2^3}{abc}=2$.
Correct Answer: 1