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


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

$I f a^{2} - b c, b^{2} - a c, c^{2} - a b i s A P w h e r e a + c$ $= 12, f i n d t h e v a l u e o f a + b + c$
	Answered by bemath last updated on 16/Jul/20
	$⇒AP : 2(b^2 −ac)=a^2 −bc+c^2 −ab 2b^2 −2ac = a^2 +c^2 −ab−bc 2b^2 −2ac = (a+c)^2 −2ac−ab−bc 2b^2 = 144−ab−b(12−a) 2b^2 = 144−12b b^2 +6b−72 = 0 ⇒(b+12)(b−6)=0→ { ((b=6)),((b=−12)) :} ∴ a+b+c = 18 or zero$
	$\Rightarrow A P : 2 (b^{2} - a c) = a^{2} - b c + c^{2} - a b$ $2 b^{2} - 2 a c = a^{2} + c^{2} - a b - b c$ $2 b^{2} - 2 a c = {(a + c)}^{2} - 2 a c - a b - b c$ $2 b^{2} = 144 - a b - b (12 - a)$ $2 b^{2} = 144 - 12 b$ $b^{2} + 6 b - 72 = 0$ $\Rightarrow (b + 12) (b - 6) = 0 \to {\begin{cases} b = 6 \\ b = - 12 \end{cases}$ $∴ a + b + c = 18 o r z e r o$
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