Let A=a2b+ab2−a2c−ac2,B=b2c+bc2−a2b−ab2 and C=a2c+ac2−b2c−bc2, where a>b>c>0. If the equation Ax2+Bx+C=0 has equal roots, then a,b,c are in
A
A.P.
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B
G.P.
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C
H.P.
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D
A.G.P.
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Solution
The correct option is C H.P. Given : A=a2b+ab2−a2c−ac2,B=b2c+bc2−a2b−ab2 and C=a2c+ac2−b2c−bc2 By observation, we see A+B+C=0 Now, the given equation is Ax2+Bx+C=0 So, one root of the equation is 1, as both roots are equal so both root are equal to 1, Product of roots CA=1⇒A=C⇒a2b+ab2−a2c−ac2=a2c+ac2−b2c−bc2⇒a[a(b−c)+(b2−c2)]=c[c(a−b)+(a2−b2)]⇒a(b−c)(a+b+c)=c(a−b)(a+b+c)⇒ab−ac=ac−bc⇒b(a+c)=2ac∴b=2aca+c