- Home
- Standard 11
- Mathematics
8. Sequences and Series
hard
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
A
$x = \frac{8}{5}$
B
$x = \frac{5}{4}$
C
$x = 2/3$
D
None of these
Solution
(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 $x = \frac{2}{3}$.
Standard 11
Mathematics
