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Question

If a(1b+1c),b(1c+1a),c(1a+1b) are in A.P., then a,b,c are also

A
A.P.
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B
G.P.
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C
H.P.
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D
None of these
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Solution

The correct option is B A.P.
If a,b,c are in A.P then

2b=a+c___________ (1)

2b(1c+1a)=a(1b+1c)+(1a+1b)c

2b(a+c)ac=a(b+c)bc+(b+a)abc

Multiplying by a,b,c

2b2(a+c)=a2(b+c)+c2(a+b)

If a,b,c are in A.P, then it should satisfy above equation
LHS=2b2(a+c)

=2b2(2b)=4b3 [ from (i)]

RHS=a2(b+c)+c2(a+b)

=a2(a+c2+c)+c2(a+a+c2)

=a2(a+3c2)+c2(3a+c2)

=a3+3a2c+3c2a+c32

=(a+c)32

=(2b)32

=4b3 Thus a,b,c are in A.P

=LHS

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