Question Number 103620 by bobhans last updated on 16/Jul/20
$${If}\:{a}^{\mathrm{2}} −{bc},\:{b}^{\mathrm{2}} −{ac},\:{c}^{\mathrm{2}} −{ab}\:{is}\:{AP}\:{where}\:{a}+{c} \\ $$$$=\:\mathrm{12},\:{find}\:{the}\:{value}\:{of}\:{a}+{b}+{c}\: \\ $$
Answered by bemath last updated on 16/Jul/20
$$\Rightarrow{AP}\::\:\mathrm{2}\left({b}^{\mathrm{2}} −{ac}\right)={a}^{\mathrm{2}} −{bc}+{c}^{\mathrm{2}} −{ab} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{ac}\:=\:{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{ac}\:=\:\left({a}+{c}\right)^{\mathrm{2}} −\mathrm{2}{ac}−{ab}−{bc} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} \:=\:\mathrm{144}−{ab}−{b}\left(\mathrm{12}−{a}\right) \\ $$$$\mathrm{2}{b}^{\mathrm{2}} \:=\:\mathrm{144}−\mathrm{12}{b} \\ $$$${b}^{\mathrm{2}} +\mathrm{6}{b}−\mathrm{72}\:=\:\mathrm{0} \\ $$$$\Rightarrow\left({b}+\mathrm{12}\right)\left({b}−\mathrm{6}\right)=\mathrm{0}\rightarrow\begin{cases}{{b}=\mathrm{6}}\\{{b}=−\mathrm{12}}\end{cases} \\ $$$$\therefore\:{a}+{b}+{c}\:=\:\mathrm{18}\:{or}\:{zero}\: \\ $$