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

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Question

(c) Let $a$ be the first term and $x$ be the common difference of the $A.P. $

Then $a + 5x = 2$

$ \Rightarrow $$a = 2 - 5x$

Let $P = {a_1}{a_

Vedclass · Accepted Answer

$x = 2/3$

Vedclass · Answer

(c) Let $a$ be the first term and $x$ be the common difference of the $A.P. $

Then $a + 5x = 2$

$ \Rightarrow $$a = 2 - 5x$

Let $P = {a_1}{a_4}{a_5} = a\,(a + 3x)\,(a + 4x)$

$ = (2 - 5x)(2 - 2x)(2 - x) = 2( - 5{x^3} + 17{x^2} - 16x + 4)$

Now $\frac{{dP}}{{dx}} = 0$

$ \Rightarrow $$x = \frac{8}{5},\;\frac{2}{3}$.

Clearly, $\frac{{{d^2}P}}{{d{x^2}}} > 0$ for $x = \frac{2}{3}$

Hence $P$ is least for $x = \frac{2}{3}$.

The sixth term of an $A.P.$ is equal to $2$, the value of the common difference of the $A.P.$ which makes the product ${a_1}{a_4}{a_5}$ least is given by

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