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Question

# If a,b,c are in A.P. and a2,b2,c2 are in G.P. such that a<b<c and a+b+c=34, then the value of a is :

A
14
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B
14122
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C
14+122
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D
14142
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Solution

## The correct option is B 14−12√2a,b,c are in A.P. ∴2b=a+c ...(1) a2,b2,c2 are in G.P. ∴(b2)2=a2c2 ...(2) a+b+c=34 ...(3) From equation (1) and (3) ⇒2b+b=34 ⇒b=14 From equation (2) ⇒a2c2=(116)2 ⇒ac=±116 ⇒a(12−a)=±116 (∵From equation (1), a+c=12) Case 1: a(12−a)=+116 ⇒2a2−a+18=0 ⇒a=1±√1−14=14 Case 2: a(12−a)=−116 ⇒2a2−a−18=0 ⇒a=1±√1+14=1±√24 ⇒a=14+12√2, 14−12√2 ∴a=14−12√2 (∵a<b<c)

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