Class 11

Math

Algebra

Sequences and Series

Let p, q, r $ϵR_{+}$ and $27pqr$ $≥(p+q+r)_{3}$ and $3p+4q+5r=12$ then $p_{3}+q_{4}+r_{5}$ is equal to

- 2
- 6
- 3
- None of these

$AM≥GM$

$3p+q+r ≥(pqr)_{31}$

$27(p+q+r)_{3} ≥pqr$

Or

$(p+q+r)_{3}≥27pqr$. ...(i)

But it is given that $27pqr≥(p+q+r)_{3}$ ...(ii)

Hence from (i) and (ii), we get

$(p+q+r)_{3}=27pqr$

Then $p=q=r=1$.

Hence $p_{3}+q_{4}+r_{5}$

$=1+1+1$

$=3$.