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Question

Suppose 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

A
12+12
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B
1212
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C
12
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D
12
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Solution

The correct option is A 1212
Let a=m-d, b=m and c=m+d where d is the common difference.
Now according to the question a,b,c are in AP
a+c=2b
Given that,
a+b+c=32
3b=32
b=12 or m=12
Also,
a2,b2,c2 are in GP . a2c2=(b2)2(md)2(m+d)2=m4(m2d2)2=m4m4+d42m2d2=m2d2(d22m2)=0d2=2m2 or d=12
Therefore a=mda=1212

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