Question

The sixth term of an A.P. $a_1,a_2,a_3,\dots ...a_n$ is 2 . If the quantity $a_1{a}_4{a}_5$ is minimum then the common difference of the A.P., is

• A. $\frac{2}{3}$
• B. $\frac{8}{5}$
• C. $\frac{1}{3}$
• D. $\frac{2}{9}$

Answer 1

Answer: (A)

$a_6=a_1+5d=2$
$\text{ Now }P=a_1{a}_4{a}_5=a_1\left(a_1+3d\right)\left(a_1+4d\right)=(2-5d)(2-2d)(2-d)=2\left(4-16d+17d^2-5d^3\right)$
$\text{ Consider }f(d)=-5d^3+17d^2-16d+4\text{ [12th, 29-08-2010 P-1] }$
$\therefore f^{\prime }(d)=-15d^2+34d-16\text{, for minima, }{f}^{\prime }(d)=0\Rightarrow d=\frac{2}{3},\frac{8}{5}$
$f^{\prime \prime }(d)=-30d+34$
$f^{\prime \prime }\left(\frac{2}{3}\right)=-20+34>0\Rightarrow d=\frac{2}{3}\text{ gives minima}$