First slide

Arithmetic progression

Question

If three positive real numbers a, b, c are in A.P. such that abc = 4, then the minimum value of b is

Moderate

A

$2^{1 / 3}$
B

$2^{2 / 3}$
C

$2^{1 / 2}$
D

$2^{3 / 2}$

Solution

Since a,b, c are in A.P., therefore, b -a = d and c -b = d,
where dis the common difference of the A.P.
$∴ a = b - d and c = b + d$
Now, abc=4
$\begin{array}{l} \Rightarrow (b - d) b (b + d) = 4 \\ \Rightarrow b (b^{2} - d^{2}) = 4 \end{array}$
But, $b (b^{2} - d^{2}) < b \times b^{2}$
$\begin{array}{l} \Rightarrow b (b^{2} - d^{2}) < b^{3} \\ \Rightarrow 4 < b^{3} \\ \Rightarrow b^{3} > 4 \\ \Rightarrow b > 2^{2 / 3} \end{array}$
Hence, the minimum value of b is 2 ^2/3

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