Class 11

Math

Algebra

Sequences and Series

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

- $2$
- $4_{31}$
- $4_{32}$
- $4$

As $a,b,c$ are in A.P.....given

$∴$ $b=2a+c $

$⇒3a+b+c ≥(abc)_{31}$

$⇒$Since $a,b,c$ are in A.P., we have $3a+b+c =b$

$⇒32b+b ≥(abc)_{31}$

$⇒b≥(abc)_{31}$

$∴b≥2$ since $abc=8$