Key Concepts and Formulas

• Arithmetic Progression (A.P.): An A.P. is a sequence where the difference between consecutive terms is constant. The $n$-th term is given by $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.
• Optimization using Calculus: To find the maximum or minimum of a function $f(x)$, we find its critical points by setting the first derivative $f'(x) = 0$. The nature of these critical points (maximum, minimum, or inflection) is determined using the second derivative test: if $f''(x) < 0$, it's a local maximum; if $f''(x) > 0$, it's a local minimum.

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Step-by-Step Solution

Step 1: Express all terms in the product in terms of a single variable. We are given an A.P. where $a_6 = 2$. Let the first term be $a_1$ and the common difference be $d$. The formula for the $n$-th term of an A.P. is $a_n = a_1 + (n-1)d$. We are given $a_6 = a_1 + (6-1)d = a_1 + 5d = 2$. From this equation, we can express $a_1$ in terms of $d$: $a_1 = 2 - 5d$ Now, we express the terms $a_1, a_4,$ and $a_5$ in terms of $d$:

• $a_1 = 2 - 5d$
• $a_4 = a_1 + (4-1)d = a_1 + 3d = (2 - 5d) + 3d = 2 - 2d$
• $a_5 = a_1 + (5-1)d = a_1 + 4d = (2 - 5d) + 4d = 2 - d$

Step 2: Formulate the product as a function of the common difference. We want to maximize the product $P = a_1 a_4 a_5$. Substituting the expressions from Step 1, we get $P$ as a function of $d$: $P(d) = (2 - 5d)(2 - 2d)(2 - d)$ To simplify the differentiation process, we expand this expression: First, multiply the last two factors: $(2 - 2d)(2 - d) = 4 - 2d - 4d + 2d^2 = 2d^2 - 6d + 4$ Now, multiply this by the first factor: $P(d) = (2 - 5d)(2d^2 - 6d + 4)$ $P(d) = 2(2d^2 - 6d + 4) - 5d(2d^2 - 6d + 4)$ $P(d) = (4d^2 - 12d + 8) - (10d^3 - 30d^2 + 20d)$ $P(d) = -10d^3 + (4d^2 + 30d^2) + (-12d - 20d) + 8$ $P(d) = -10d^3 + 34d^2 - 32d + 8$

Step 3: Find the critical points by differentiating and setting the first derivative to zero. To find the values of $d$ that maximize or minimize $P(d)$, we compute the first derivative with respect to $d$: $P'(d) = \frac{d}{dd}(-10d^3 + 34d^2 - 32d + 8)$ $P'(d) = -30d^2 + 68d - 32$ Set $P'(d) = 0$ to find the critical points: $-30d^2 + 68d - 32 = 0$ Divide by $-2$ to simplify the quadratic equation: $15d^2 - 34d + 16 = 0$ We solve this quadratic equation using the quadratic formula $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=15$, $b=-34$, and $c=16$: $d = \frac{-(-34) \pm \sqrt{(-34)^2 - 4(15)(16)}}{2(15)}$ $d = \frac{34 \pm \sqrt{1156 - 960}}{30}$ $d = \frac{34 \pm \sqrt{196}}{30}$ $d = \frac{34 \pm 14}{30}$ This gives two critical values for $d$: $d_1 = \frac{34 + 14}{30} = \frac{48}{30} = \frac{8}{5}$ $d_2 = \frac{34 - 14}{30} = \frac{20}{30} = \frac{2}{3}$

Step 4: Use the second derivative test to determine which critical point yields a maximum. We compute the second derivative of $P(d)$: $P''(d) = \frac{d}{dd}(-30d^2 + 68d - 32)$ $P''(d) = -60d + 68$ Now, we evaluate $P''(d)$ at each critical point:

• For $d = \frac{8}{5}$: $P''\left(\frac{8}{5}\right) = -60\left(\frac{8}{5}\right) + 68 = -12 \times 8 + 68 = -96 + 68 = -28$ Since $P''\left(\frac{8}{5}\right) < 0$, this critical point corresponds to a local maximum.
• For $d = \frac{2}{3}$: $P''\left(\frac{2}{3}\right) = -60\left(\frac{2}{3}\right) + 68 = -20 \times 2 + 68 = -40 + 68 = 28$ Since $P''\left(\frac{2}{3}\right) > 0$, this critical point corresponds to a local minimum.

Thus, the common difference that maximizes the product $a_1 a_4 a_5$ is $\frac{8}{5}$.

Common Mistakes & Tips

• Algebraic Errors: Carefully check all algebraic expansions and simplifications, as mistakes here are common and can lead to incorrect derivatives and solutions.
• Correct Application of Calculus: Ensure the first derivative is set to zero to find critical points and the second derivative is used correctly to classify these points as maxima or minima.
• Interpreting the Question: Double-check that you are indeed finding the value that maximizes the product and not just any critical point.

Summary To maximize the product $a_1 a_4 a_5$ for an A.P. with $a_6 = 2$, we first expressed all terms in the product as a function of the common difference $d$. This resulted in a cubic polynomial $P(d)$. By finding the critical points of $P(d)$ using the first derivative and then classifying them with the second derivative test, we identified that the common difference $d = \frac{8}{5}$ yields the maximum product.

The final answer is $\boxed{8 \over 5}$.