Question

Let \( a, a + r \) and \( a + 2r \) be positive real numbers such that their product is 64. Then the minimum value of \( a + 2r \) is equal to:

• A. \( 4 \)
• B. \( 3 \)
• C. \( 2 \)
• D. \( \frac{1}{2} \)
• E. \( 1 \)

Hint:

For positive numbers in A.P. with a fixed product, the individual terms are at their "least extreme" values when the common difference \( r \) is zero.

Answer 1

The Correct Option is A

Concept: The problem involves numbers in an Arithmetic Progression (A.P.). According to the Arithmetic Mean-Geometric Mean (AM-GM) inequality, for any set of positive real numbers, the Arithmetic Mean is always greater than or equal to the Geometric Mean. The equality holds only when all the numbers are equal.

Step 1: Apply AM-GM to the three terms.
Let the three positive numbers be \( x = a \), \( y = a+r \), and \( z = a+2r \). Given their product: \( x \cdot y \cdot z = 64 \). By AM-GM inequality: \[ \frac{x + y + z}{3} \geq \sqrt[3]{xyz} \] \[ \frac{a + (a+r) + (a+2r)}{3} \geq \sqrt[3]{64} \] \[ \frac{3a + 3r}{3} \geq 4 \implies a + r \geq 4 \]

Step 2: Identify the condition for the minimum value.
The value of \( a+2r \) is minimized when the terms are as close together as possible. In the case of a constant product, the terms themselves are minimized when they are all equal (i.e., when \( r = 0 \)). If \( a = a+r = a+2r \), then: \[ a \cdot a \cdot a = 64 \implies a^3 = 64 \implies a = 4 \] In this scenario, \( a + 2r = 4 + 2(0) = 4 \). If \( r > 0 \), then \( a + 2r \) would increase to maintain the product of 64 while \( a \) decreases. Therefore, the minimum value is 4.