JEE PYQ: Sequence And Series Question 55
Question 55 - 2019 (10 Apr Shift 2)
Let $a_1, a_2, a_3, \ldots$ be an A.P. with $a_6 = 2$. Then the common difference of this A.P., which maximises the product $a_1 a_4 a_5$ is:
(1) $\frac{3}{2}$
(2) $\frac{8}{5}$
(3) $\frac{6}{5}$
(4) $\frac{2}{3}$
Show Answer
Answer: (2)
Solution
Here, $a$ is first term of A.P. and $d$ is common difference
Let $A = a_1 a_4 a_5 = a(a + 3d)(a + 4d)$
$A = (2 - 5d)(2 - 2d)(2 - d)$
By $\frac{dA}{dd} = 0$
$\Rightarrow (2 - 5d)(-6 + 6d + 2d^2)(-5) = 0$
$\Rightarrow -15d^2 + 34d - 16 = 0 \Rightarrow d = \frac{8}{5}, \frac{2}{3}$
For $d = \frac{8}{5}$, $\frac{d^2A}{dd^2} < 0$. Hence $d = \frac{8}{5}$
BETA
