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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)

best_answer

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

Posted by

Himanshu

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