# 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) Option 2) Option 3) Option 4)

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)

Correct option

Option 2)

Incorrect option

Option 3)

Incorrect option

Option 4)

Incorrect option

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