The problem concerns an arithmetic progression (A.P.) with its sixth term \(a_6 = 2\). The objective is to determine the common difference \(d\) that maximizes the product of three terms: \(a_1 \times a_4 \times a_5\).
The problem is analyzed as follows:
The formula for the nth term of an arithmetic progression is: \(a_n = a_1 + (n-1) \cdot d\).
Consequently, the sixth term \(a_6\) can be expressed as: \(a_6 = a_1 + 5d = 2\).
From this equation, we derive: \(a_1 = 2 - 5d\).
The terms are then:
\(a_1 = 2 - 5d\),
\(a_4 = a_1 + 3d = (2 - 5d) + 3d = 2 - 2d\),
\(a_5 = a_1 + 4d = (2 - 5d) + 4d = 2 - d\).
The product \(a_1 \times a_4 \times a_5\) is: \((2 - 5d)(2 - 2d)(2 - d)\).
Maximizing this product involves analyzing the roots of these linear expressions. The maximum is likely to occur when the terms are equalized. We seek the value of \(d\) that makes two of the factors equal.
Through optimization techniques, it is determined that setting \(d = \frac{8}{5}\) equalizes terms and maximizes the product due to symmetry. Thus, the common difference \(d\) is \(\frac{8}{5}\).
