Question

We are giving the concept of A.M of $m^{th}$ power.Let $a,b>0$ and $a\ne b$ and let $m$ be a real number. Then
$\frac{a^m+b^m}{2}>{\left(\frac{a+b}{2}\right)}^m$ if $m\in R-[0,1]$
However, if $m\in (0,1)$ then $\frac{a^m+b^m}{2}<{\left(\frac{a+b}{2}\right)}^m$.
Obviously, if $m\in \{0,1\}$ then $\frac{a^m+b^m}{2}={\left(\frac{a+b}{2}\right)}^m$.
On the basis of the above information, answer the following questions:

If $a,b,c$ be positive and in H.P and if $\lambda =\frac{a^n+c^n}{b^n}$ for all $n\notin (0,1),$ then the correct statement is:

• A. $\lambda <2$
• B. $\lambda >2$
• C. $\lambda =2$
• D. none of these

Answer 1

The correct option is B $\lambda >2$

By using the concept of A.M we have

$\frac{a^n+c^n}{2}>{\left(\frac{a+c}{2}\right)}^n$

or $a^n+c^n>2{\left(\frac{a+c}{2}\right)}^n$ .........$\left(1\right)$

But we know that A.M$>$G.M$>$H.M.

$\therefore$A.M$>$H.M

$\Rightarrow \frac{a+c}{2}>b$

or ${\left(\frac{a+c}{2}\right)}^n>b^n$

$\Rightarrow 2{\left(\frac{a+c}{2}\right)}^n>2b^n$ ......$\left(2\right)$

From eqns$\left(1\right)$ and $\left(2\right)$, we get

$a^n+c^n>2b^n$

or $\frac{a^n+c^n}{b^n}>2$

Thus $\lambda >2$