If three positive numbers a, b and c are in A.P. such that abc=8, then the minimum possible value of b is :

  • Option 1)

    2

  • Option 2)

    4^{\frac{1}{3}}

  • Option 3)

    4^{\frac{2}{3}}

  • Option 4)

    4

 

Answers (2)
N neha
H Himanshu

Use the Concept of

Arithmetic mean of two numbers (AM) -

A=\frac{a+b}{2}

- wherein

It is to be noted that the sequence a, A, b, is in AP where, a and b are the two numbers.

 

 a,b,c in A.P so that 

b=\frac{a+c}{2}

2b=a+c

\therefore\: c=2b-a

=> abc=8

=>  ab\left [ 2b-a \right ]=8

\therefore \: 2ab^{2} - a^{2}b=8=K

\therefore\: 2+1\times b^{2}-2a\times b=0

\therefore\: 2b\left [ b-a \right ]=0

      b\neq 0

     but  b=a

     then  c=a

 \therefore \: a^{3} =8

   a=2=b=c   

\therefore \: a=b=c=8 

 


Option 1)

2

Correct option    

Option 2)

4^{\frac{1}{3}}

Incorrect option    

Option 3)

4^{\frac{2}{3}}

Incorrect option    

Option 4)

4

Incorrect option    

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