Question

The $6^{th}$ term of an AP is equal to 2 value of common difference of the AP which makes the product $a$ , $a_4$ , $a_5$ least is given by

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

$a_6=2$

$a+5d=2$

$a\times a_4\times a_5=$ least

$\begin{array}{cc}=a(a+3d)(a+4d)& \\ \text{put}a=2-5d& \end{array}$

$=(2-5d)(2-5d+3d)(2-5d+4d)$

$=(2-5d)(2-2d)(2-d)$

$=(2-5d)(4-6d+2d^2)$

$=8-12d+4d^2-20d+30d^2-{100}^3$

$=-10d^3+34d^2-32d+8$

On differentiating and equate to zero

$-30d^2+68d-32=0$

$-15d^2+34d-16=0$

$-15d^2+10d+24d-16=0$

$5d(-3d+2)-8(-3d+2)=0$

$(5d-8)(2-3d)=0$

$d=\frac{8}{5}$ or $d=\frac{2}{3}$