Question Number 103620 by bobhans last updated on 16/Jul/20

If a^2 −bc, b^2 −ac, c^2 −ab is AP where a+c  = 12, find the value of a+b+c

Answered by bemath last updated on 16/Jul/20

⇒AP : 2(b^2 −ac)=a^2 −bc+c^2 −ab  2b^2 −2ac = a^2 +c^2 −ab−bc  2b^2 −2ac = (a+c)^2 −2ac−ab−bc  2b^2  = 144−ab−b(12−a)  2b^2  = 144−12b  b^2 +6b−72 = 0  ⇒(b+12)(b−6)=0→ { ((b=6)),((b=−12)) :}  ∴ a+b+c = 18 or zero