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Question

Let a,b,c be in A.P. and let a2,b2,c2 are in G.P. If a<b<c and a+b+c=32, then a=

A
122
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B
123
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C
12
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D
12 12
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Solution

The correct option is D 12 12

If a,b,c are in AP then,

2b=a+c

b=12(a+c) … (i)

If a2,b2,c2 are in GP then b4=a2c2 or b2=±ac … (ii)

Using (i) in a+b+c=32 gives 3b=32

b=12

Hence a+c=1 & ac=±14

So a & c are roots of either x2x+14=0 or x2x14=0

The first has equal roots of x=12 and second gives x==121±2 for a & c

Since a<c we must have x=12(12)

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