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If three positive real numbers are in A.P. such that , then the minimum possible value of is
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a
23/2
b
22/3
c
21/3
d
25/2
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Correct option is B
Let d be the common difference of the A.P., then 4=abc=(b−d)b(b+d)=bb2−d2⇒ b3=4+bd2≥4 ∵b>0,d2≥0⇒ b≥22/3 Thus, minimum possible value of b is 22/3, that is the case when d=0.Similar Questions
If and 1 are in A.P, then x is equal to
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