The sixth term of an A.P. is equal to 2. The | vedclass

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(C) Let $a$ be the first term and $x$ be the common difference of the $A.P.$Given that the sixth term $a_6 = a + 5x = 2$,so $a = 2 - 5x$.Let the produ

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$x = \frac{2}{3}$

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(C) Let $a$ be the first term and $x$ be the common difference of the $A.P.$Given that the sixth term $a_6 = a + 5x = 2$,so $a = 2 - 5x$.Let the product $P = a_1 a_4 a_5 = a(a + 3x)(a + 4x)$.Substituting $a = 2 - 5x$:$P = (2 - 5x)(2 - 5x + 3x)(2 - 5x + 4x) = (2 - 5x)(2 - 2x)(2 - x)$.$P = (2 - 5x)(4 - 6x + 2x^2) = 8 - 12x + 4x^2 - 20x + 30x^2 - 10x^3 = -10x^3 + 34x^2 - 32x + 8$.To find the minimum,set $\frac{dP}{dx} = -30x^2 + 68x - 32 = 0$.Dividing by $-2$: $15x^2 - 34x + 16 = 0$.Using the quadratic formula: $x = \frac{34 \pm \sqrt{34^2 - 4(15)(16)}}{2(15)} = \frac{34 \pm \sqrt{1156 - 960}}{30} = \frac{34 \pm \sqrt{196}}{30} = \frac{34 \pm 14}{30}$.So,$x = \frac{48}{30} = \frac{8}{5}$ or $x = \frac{20}{30} = \frac{2}{3}$.Checking the second derivative $\frac{d^2P}{dx^2} = -60x + 68$.For $x = \frac{8}{5}$,$\frac{d^2P}{dx^2} = -60(\frac{8}{5}) + 68 = -96 + 68 = -28 For $x = \frac{2}{3}$,$\frac{d^2P}{dx^2} = -60(\frac{2}{3}) + 68 = -40 + 68 = 28 > 0$ (Local Minimum).Thus,the product is least for $x = \frac{2}{3}$.

The sixth term of an $A.P.$ is equal to $2$. The value of the common difference $x$ of the $A.P.$ that makes the product $a_1 a_4 a_5$ least is given by:

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