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 $a = 2 - 5x$ into the product:$P = (2 - 5x)(2 - 5x + 3x)(2 - 5x + 4x)$$P = (2 - 5x)(2 - 2x)(2 - x)$$P = (2 - 5x)(4 - 6x + 2x^2) = 8 - 12x + 4x^2 - 20x + 30x^2 - 10x^3$$P = -10x^3 + 34x^2 - 32x + 8 = 2(-5x^3 + 17x^2 - 16x + 4)$.To find the minimum,differentiate $P$ with respect to $x$:$\frac{dP}{dx} = 2(-15x^2 + 34x - 16) = 0$.Solving $-15x^2 + 34x - 16 = 0$ or $15x^2 - 34x + 16 = 0$: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 $\frac{d^2P}{dx^2} = 2(-30x + 34)$:For $x = \frac{8}{5}$,$\frac{d^2P}{dx^2} = 2(-48 + 34) For $x = \frac{2}{3}$,$\frac{d^2P}{dx^2} = 2(-20 + 34) > 0$ (Local minimum).Thus,$P$ 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 $A.P.$ which makes the product $a_1 a_4 a_5$ least is given by:

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