In an A.P.,the sixth term is a_6 = 2. If the | vedclass

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(C) Given $a_6 = 2$,we have $a + 5d = 2$,which implies $a = 2 - 5d$.Let the product be $P = a_1 a_4 a_5 = a(a + 3d)(a + 4d)$.Substituting $a = 2 - 5d$

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$\frac{8}{5}$

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(C) Given $a_6 = 2$,we have $a + 5d = 2$,which implies $a = 2 - 5d$.Let the product be $P = a_1 a_4 a_5 = a(a + 3d)(a + 4d)$.Substituting $a = 2 - 5d$:$P = (2 - 5d)(2 - 5d + 3d)(2 - 5d + 4d)$$P = (2 - 5d)(2 - 2d)(2 - d)$$P = (2 - 5d)(4 - 2d - 4d + 2d^2) = (2 - 5d)(4 - 6d + 2d^2)$$P = 8 - 12d + 4d^2 - 20d + 30d^2 - 10d^3$$P(d) = -10d^3 + 34d^2 - 32d + 8$To find the maximum,we find the derivative $P'(d)$:$P'(d) = -30d^2 + 68d - 32$Set $P'(d) = 0$:$-2(15d^2 - 34d + 16) = 0$$-2(15d^2 - 24d - 10d + 16) = 0$$-2[3d(5d - 8) - 2(5d - 8)] = 0$$-2(3d - 2)(5d - 8) = 0$Critical points are $d = \frac{2}{3}$ and $d = \frac{8}{5}$.Using the second derivative test or sign scheme of $P'(d)$:$P''(d) = -60d + 68$.For $d = \frac{2}{3}$,$P''(\frac{2}{3}) = -60(\frac{2}{3}) + 68 = -40 + 68 = 28 > 0$ (Local Minima).For $d = \frac{8}{5}$,$P''(\frac{8}{5}) = -60(\frac{8}{5}) + 68 = -96 + 68 = -28 Thus,the product is greatest when $d = \frac{8}{5}$.

In an $A.P.$,the sixth term is $a_6 = 2$. If the product $a_1 a_4 a_5$ is the greatest,then the common difference of the $A.P.$ is equal to

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