Question

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 $a_1{a}_4{a}_5$ greatest, is

• A. $\frac{8}{5}$
• B. $\frac{2}{3}$
• C. $\frac{3}{5}$
• D. $\frac{3}{4}$

Answer 1

Answer: (A)

Given: $a_1+5d=2$
Let $y=a_1{a}_4{a}_5=a_1\left(a_1+3d\right)\left(a_1+4d\right)$
$=(2-5d)(2-2d)(2-d)$
$($ Putting $a_1=2-5d)$
$=2\left(4-16d+17d^2-5d^3\right)$
The value of $d$ at which $y$ attains maxima is given by
$\frac{dy}{dx}=0$(by calculus)
$\Rightarrow -16+34d-15d^2=0$
$\Rightarrow d=\frac{34\pm 14}{30}=\frac{2}{3},\frac{8}{5}$
Now, $\frac{d^{2}y}{d{x}^2}=34-30d>0\text{ for }d=\frac{2}{3}$
$<0$ for $d=\frac{8}{5}$
Hence, $y$ is maximum for $d=\frac{8}{5}$