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Question

If a,b,c are in A.P and if (b−c)x2+(c−a)+a−b=0 and 2(c+a)x2+(b+c)x=0 have a common root then

A
a2,b2,c2 are in A.P
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B
a2,c2,b2 are in A.P
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C
a2,c2,b2 are in G.P
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D
a2,b2,c2 are in H.P
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Solution

The correct option is A a2,b2,c2 are in A.P
Given a,b,c are in AP

2b=a+c

(bc)x2+(ca)x+(ab)=0

(bc)x2(bc)x(ab)x+(ab)=0

(bc)x(x1)(ab)(x1)=0

(x1)[(bc)x(ab)]=0

x=abbc,1

x=abab(2b=a+cba=cb)

x=1

Now 2(c+a)x2+(b+c)x=0

x[2(c+a)x+(b+c)]=0

x=0

x=(b+c)2.2b

x=b+c4b

Now

(b+c)4b=1

4b=b+c

5b=c

a+c=2b

a5b=2b

a=7d

c2a2=25b249b2=24b2

b2c2=b225b2=24b2

So,

c2a2=b2c2

hence a2,c2,b2 are in Ap


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