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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

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a
1412
b
14132
c
14142
d
14122

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detailed solution

Correct option is D

Given that, a,b,c care in A.P. 2b=a+c(i)

Again, a2,b2,c2 are in G.P.

b4=(ac)2b22(ac)2=0b2acb2+ac=0b2=ac  or  b2=ac           …(ii)

But b2=ac is not possible.

(This gives a=b=c=1/4)

Also, a+b+c=3/43b=3/4b=1/4

Putting this value of b in (i) and (ii), we have

1/2=a+c ...(iii)   and  1/16=ac         …(iv)

From (iii), we have 1/2 a=c

Putting the value of c in (iv), we have 16a28a1=0

a=8±64+642×16=8±822×16a=1±24=14±122a=14122

 

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