Let a_1, a_2, a_3, dots be an A.P. with a_6 = | vedclass

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(B) Let $a$ be the first term and $d$ be the common difference of the $A.P.$Given $a_6 = a + 5d = 2$,so $a = 2 - 5d$.The terms are $a_1 = a = 2 - 5d$,

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$\frac{8}{5}$

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(B) Let $a$ be the first term and $d$ be the common difference of the $A.P.$Given $a_6 = a + 5d = 2$,so $a = 2 - 5d$.The terms are $a_1 = a = 2 - 5d$,$a_4 = a + 3d = 2 - 2d$,and $a_5 = a + 4d = 2 - d$.Let the product be $f(d) = a_1 a_4 a_5 = (2 - 5d)(2 - 2d)(2 - d)$.Expanding the expression: $f(d) = (4 - 4d - 10d + 10d^2)(2 - d) = (10d^2 - 14d + 4)(2 - d) = 20d^2 - 10d^3 - 28d + 14d^2 + 8 - 4d = -10d^3 + 34d^2 - 32d + 8$.To find the maximum,find the derivative $f'(d) = -30d^2 + 68d - 32$.Set $f'(d) = 0 \Rightarrow -2(15d^2 - 34d + 16) = 0 \Rightarrow 15d^2 - 34d + 16 = 0$.Using the quadratic formula: $d = \frac{34 \pm \sqrt{34^2 - 4(15)(16)}}{2(15)} = \frac{34 \pm \sqrt{1156 - 960}}{30} = \frac{34 \pm \sqrt{196}}{30} = \frac{34 \pm 14}{30}$.So,$d_1 = \frac{48}{30} = \frac{8}{5}$ and $d_2 = \frac{20}{30} = \frac{2}{3}$.Find the second derivative: $f''(d) = -60d + 68$.At $d = \frac{2}{3}$,$f''(\frac{2}{3}) = -60(\frac{2}{3}) + 68 = -40 + 68 = 28 > 0$ (local minimum).At $d = \frac{8}{5}$,$f''(\frac{8}{5}) = -60(\frac{8}{5}) + 68 = -96 + 68 = -28 Thus,the product is maximized at $d = \frac{8}{5}$.

Let $a_1, a_2, a_3, \dots$ be an $A.P.$ with $a_6 = 2$. Then the common difference of this $A.P.$,which maximizes the product $a_1 a_4 a_5$,is

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