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01 Nov 2012, 19:12
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If the roots of a(b-c)x^2 +b(c-a)x + c(a-b) are equal, then a,b,c are in ?
A. A.P
B. H.P
C. A.G.P
D. G.P
E. None
Is A.G.P part of the gmat.. what difficulty level would you rate this questions ?
A. A.P
B. H.P
C. A.G.P
D. G.P
E. None
Is A.G.P part of the gmat.. what difficulty level would you rate this questions ?
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02 Nov 2012, 01:55
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chris93
Two points:
1) The question is incomplete, roots are of an equation, this has no RHS. ( = 0 part is missing)
question should be modified as :
If the roots of a(b-c)x^2 +b(c-a)x + c(a-b) =0 are equal, then a,b,c are in ?
2) I dont think this could be part of GMAT. Probably only at a difficulty level of 800+
However, having said that, lets look at solving such problem:
First of all, important concept:
if in equation Ax^2+Bx+C=0 roots are equal that would mean B^2 -4AC =0
or B^2 =4AC .
Applying this in the given equation a(b-c)x^2 +b(c-a)x + c(a-b)=0 we get:
\([b(c-a)]^2 = 4 * a(b-c) *c(a-b)\)
So we need to work with this equation.
Here you can observe first -
if
a b c were in gp then b^2 =ac or b/a = c/b = sqrt(c/a)
and
if a b c were in AP then 2b = a+c or b-a = c-b = (c-a)/2
Also, for AGP, terms are multiples of AP and GP.
To start with I'm not considering HP as I dont see any term in the form of a+b or b+c or c+a etc.
Lets, rewrite, our original equation in a form we can identify our sequences.
\(b^2 * (c-a)^2 = ac * 2(b-c) * 2(a-b)\)
or
\(b^2 *(c-a)* (c-a) = ac * 2(b-c) * 2(a-b)\)
If you notice, terms in LHS and RHS are basically multiplication of AP and GP. Thus a, b and c are in A.G.P.
Ans C it is.
Hope it helps
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