Question:

Step 1: Set up using the constraint $a_6 = 2$ Let the first term of the arithmetic progression be $a$ and the common difference be $d$. The $n$-th term of an AP is given by $a_n = a + (n-1)d$. Given that the sixth term $a_6 = 2$: $$a + 5d = 2$$ Thus, the first term can be expressed as $a = 2 - 5d$. Step 2: Express the required terms in terms of $d$ The terms $a_1, a_4, a_5$ are: $$a_1 = a = 2 - 5d$$ $$a_4 = a + 3d = (2 - 5d) + 3d = 2 - 2d$$ $$a_5 = a + 4d = (2 - 5d) + 4d = 2 - d$$ Step 3: Form the product $P$ The product $P = a_1 a_4 a_5$ is: $$P = (2 - 5d)(2 - 2d)(2 - d)$$ Step 4: Expand the product First, multiply the first two factors: $$(2 - 5d)(2 - 2d) = 4 - 4d - 10d + 10d^2 = 10d^2 - 14d + 4$$ Now, multiply this result by the third factor: $$P = (10d^2 - 14d + 4)(2 - d)$$ $$P = 20d^2 - 10d^3 - 28d + 14d^2 + 8 - 4d$$ $$P = -10d^3 + 34d^2 - 32d + 8$$ Step 5: Find the critical points by differentiation To find the value of $d$ that minimizes $P$, we differentiate $P$ with respect to $d$ and set the derivative to zero: $$\frac{dP}{dd} = \frac{d}{dd}(-10d^3 + 34d^2 - 32d + 8)$$ $$\frac{dP}{dd} = -30d^2 + 68d - 32$$ Set $\frac{dP}{dd} = 0$: $$-30d^2 + 68d - 32 = 0$$ $$30d^2 - 68d + 32 = 0$$ Dividing by 2: $$15d^2 - 34d + 16 = 0$$ Step 6: Solve for $d$ using the quadratic formula Using the quadratic formula $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for $15d^2 - 34d + 16 = 0$: $$d = \frac{34 \pm \sqrt{(-34)^2 - 4(15)(16)}}{2(15)}$$ $$d = \frac{34 \pm \sqrt{1156 - 960}}{30}$$ $$d = \frac{34 \pm \sqrt{196}}{30}$$ $$d = \frac{34 \pm 14}{30}$$ This yields two possible 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 7: Determine which value of $d$ gives the minimum product To determine whether these critical points correspond to a maximum or minimum, we use the second derivative test. The second derivative of $P$ with respect to $d$ is: $$\frac{d^2P}{dd^2} = \frac{d}{dd}(-30d^2 + 68d - 32)$$ $$\frac{d^2P}{dd^2} = -60d + 68$$ Evaluate the second derivative at each critical point: For $d = \frac{8}{5}$: $$\frac{d^2P}{dd^2} \Big|_{d=8/5} = -60\left(\frac{8}{5}\right) + 68 = -96 + 68 = -28$$ Since $\frac{d^2P}{dd^2} < 0$, $d = \frac{8}{5}$ corresponds to a local maximum. For $d = \frac{2}{3}$: $$\frac{d^2P}{dd^2} \Big|_{d=2/3} = -60\left(\frac{2}{3}\right) + 68 = -40 + 68 = 28$$ Since $\frac{d^2P}{dd^2} > 0$, $d = \frac{2}{3}$ corresponds to a local minimum. The common difference that makes the product $a_1 a_4 a_5$ least is $d = \frac{2}{3}$.