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Question

Supposse a,b,c are in A.P. and a2,b2,c2 are in G.P. If a<b<c and a+b+c=32, then the value of a is 
  1. 1213
  2. 122
  3. 123
  4. 1212


Solution

The correct option is D 1212
Given that a,b,c are in A.P.
2b=a+c
But given a+b+c=323b=32b=12 and then a+c=1
Again a2,b2,c2,are in G.P.b4=a2c2b2=±acac=14 or14
 and a+c=1     ...(i)
Considering a+c=1 and ac=14(ac)2=11=0a=c, but 
ac as given that a<b<c<
 We consider a+c=1 and ac=14(ac)2=1+1=2ac=±2
but a<cac=2     ...(ii)
Solving (i) and (ii) we get a=1212

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