Question

Let $a_1,a_2,a_3,...$ 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 :

• A. $\frac{2}{3}$
• B. $\frac{8}{5}$
• C. $\frac{6}{5}$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $\frac{8}{5}$

First term of the A.P.$=a$
Common difference $=d$
$\therefore a+5d=2$
$a_1\cdot a_4\cdot a_5=a(a+3d)(a+4d)=(2-5d)(2-2d)(2-d)=2[-5d^3+17d^2-16d+4]$

Now, assuming
$f\left(d\right)=2[-5d^3+17d^2-16d+4]\Rightarrow f^{\prime }\left(d\right)=2[-15d^2+34d-16]\Rightarrow f^{\prime }\left(d\right)=0\Rightarrow (3d-2)(5d-8)=0\Rightarrow d=\frac{2}{3},\frac{8}{5}$
Now,
$f^{\prime \prime }\left(d\right)=2[-30d+34]\Rightarrow f^{\prime \prime }\left(d\right)=-60(d-\frac{17}{15})$
At $d=\frac{2}{3}$, we get
$f^{\prime \prime }\left(d\right)>0$
At $d=\frac{8}{5}$, we get
$f^{\prime \prime }\left(d\right)<0$

Hence, the required common difference is $\frac{8}{5}$