Mathematics - Sequence & Series Question with Solution | TestHub

Let $d$ be the common difference: $d=b-a=c-b$, so $c=2b-a\dots (1)$ $(a-c{)}^2=(-2d{)}^2=4d^2=4(b-a{)}^2=4\left(b^2+a^2-2ab\right)=4\left(b^2-a(2b-a)\right)=4\left(b^2-ac\right)\text{ by (1). }$