Question

If three positive real numbers $a$, $b$ and $c$ are in A.P. such that $abc=8$, then the minimum possible value of $b$ is$A.2$
B.{{4}^{\dfrac{1}{3}}}$$$$$ C.{{4}^{\dfrac{2}{3}}}$$$$$
D.4$$$$

Answer 1

Use the general three terms of an AP and given condition to proceed. Alternatively you can use the relation between arithmetic and geometric means.$$$$

Complete step by step answer:
It is given that the three numbers $a$, $b$ and $c$ are in A.P or in arithmetic progression. Any three numbers in an A.P. are given by $x-d,x,x+d$ where $x$ is any term in a sequence and $d$ is the common difference . As $a$, $b$ and $c$ are in A.P we can assign $a=x-d,b=x,c=x+d$ and observe
$a=b-d,b,c=b+d$
Another relation is given in the question as $abc=8$. Putting the value of $c$,

& abc=8 \\\ & \Rightarrow \left( b-d \right)\left( b \right)\left( b+d \right)=8 \\\ \end{aligned}$$ Using the algebraic identify ${{a}^{2}}-{{b}^{2}}$ we get , $$\begin{aligned} & b\left( {{b}^{2}}-{{d}^{2}} \right)=8 \\\ & \Rightarrow {{2}^{3}}=b({{b}^{2}}-{{d}^{2}}) \\\ & \Rightarrow {{2}^{3}}={{b}^{3}}-b{{d}^{2}} \\\ & \Rightarrow {{b}^{3}}={{2}^{3}}+b{{d}^{2}} \\\ \end{aligned}$$ As given in the question that $a$, $b$ and $c$ are positive real numbers. So the value $b{{d}^{2}}\ge 0$ as $b$ is non-negative as given in the question and also ${{d}^{2}}$ being a square is always non-negative. So we take cube root both side and get $$\begin{aligned} & \Rightarrow {{b}^{3}}\ge {{2}^{3}} \\\ & \Rightarrow b\ge 2 \\\ \end{aligned}$$ So the minimum possible value of b is 2. **So, the correct answer is “Option A”.** **Note:** Alternative method: The arithmetic mean of three numbers is $\dfrac{a+b+c}{3}$ and the geometric mean of three numbers is given by $\sqrt[3]{abc}$ . we know from the inequality relation that arithmetic mean is always is greater than or equal to geometric mean. In symbols, $$\begin{aligned} & AM\ge GM \\\ & \Rightarrow \dfrac{a+b+c}{3}\ge \sqrt[3]{abc} \\\ \end{aligned}$$ Also $a+c=x-d+x+d=2b$, Replacing above, $$\begin{aligned} & \Rightarrow \dfrac{\left( a+c \right)+b}{3}\ge \sqrt[3]{8} \\\ & \Rightarrow \dfrac{\left( a+c \right)+b}{3}\ge 2 \\\ & \Rightarrow \dfrac{3b}{3}\ge 2 \\\ & \Rightarrow b\ge 2 \\\ \end{aligned}$$