Search any question & find its solution
Question:
Answered & Verified by Expert
If three positive real numbers $a, b, c$ are in A.P. and $a b c=4$ then minimum possible value of $b$ is
Options:
Solution:
2603 Upvotes
Verified Answer
The correct answer is:
$2^{2 / 3}$
$$
\text { Hints } \begin{aligned}
& (b-d) b(b+d)=4 \\
& \left(b^2-d^2\right) b=4 \\
& b^3=4+d^2 b \\
& b^3 \geq 4 \Rightarrow b \geq(2)^{2 / 3}
\end{aligned}
$$
\text { Hints } \begin{aligned}
& (b-d) b(b+d)=4 \\
& \left(b^2-d^2\right) b=4 \\
& b^3=4+d^2 b \\
& b^3 \geq 4 \Rightarrow b \geq(2)^{2 / 3}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.