Question

The fourth term of an arithmetic progression is $4$. At what value of the difference of the progression is the sum of the pairwise products of the first three terms of the progression the least?

Answer 1

Given $a_4=4$

$⟹a_1=4-3d,a_2=4-2d,a_3=4-d$

sum of product of these terms are $S=(4-3d)(4-2d)+(4-2d)(4-d)+(4-d)(4-3d)=11d^2-40d+48$

$\because S$ is least $d=\frac{-(-40)}{2\left(11\right)}=\frac{20}{11}$