Question

If a,b,c are distinct, positive and in HP, then ${a^n} + {c^n}$ is
1. $ > {b^n}$
2. \[ > 2{b^n}\]
3. \[ < {b^n}\]
4. \[ < 2{b^n}\]

Answer 1

Hint: This question requires the use of the inequality involving arithmetic progression, geometric progression and harmonic progression. We first calculate the arithmetic mean and geometric mean of two numbers and apply the inequality. Then, we apply the inequality for geometric and harmonic mean of different numbers. Using both the results, we get to the required result.
Formula for A.M. of two numbers x and y is $\left( {\dfrac{{x + y}}{2}} \right)$.
Formula for G.M. of two numbers x and y is $\sqrt {xy} $.
Formula for H.M. of two numbers x and y is $\dfrac{{2xy}}{{x + y}}$.

Complete step-by-step solution:
Now, we have a, b and c as distinct numbers.
Let’s take \[{a^n}\]and \[{c^n}\] as two numbers and apply the A.M. and G.M. inequality i.e., \[A.M. > G.M\] for distinct positive numbers.
So, we get,
\[ \Rightarrow \dfrac{{{a^n} + {c^n}}}{2} > \sqrt {{{\left( {ac} \right)}^n}} \]
Cross multiplying the terms, we get,
\[ \Rightarrow {a^n} + {c^n} > 2\sqrt {{{\left( {ac} \right)}^n}} - - - - \left( 1 \right)\]
Now, let’s take a and c as our numbers and apply the H.M. and G.M. inequality i.e., \[G.M. > H.M\] for distinct positive numbers, we get,
\[ \Rightarrow \sqrt {(ac)} > b\]
And now raising both the sides with a power of n,
\[ \Rightarrow {(ac)^{\dfrac{n}{2}}} > {b^n} - - - - \left( 2 \right)\]
Now by using \[(1)\] and \[(2)\], we have,
\[ \Rightarrow \dfrac{{{a^n} + {c^n}}}{2} > {b^n}\]
Cross multiplying the terms in equation, we get,
\[ \Rightarrow {a^n} + {c^n} > 2{b^n}\]
Hence, option (B) is the correct answer.

Note: Conceptual mistakes related to the A.M.-G.M.-H.M. inequality can occur. One should be well versed with those concepts. Take care of the calculations while doing such questions. The sign of the inequality remains the same when a positive number is multiplied on both sides. Whereas the sign of inequality changes when multiplied with a negative number.