Question:
If \(a, b\) and \(c\) are in HP, then for any \(n \in \mathbb{N}\), which one of the following is true?
If \(a, b\) and \(c\) are in HP, then for any \(n \in \mathbb{N}\), which one of the following is true?
Show Hint
HP $\Rightarrow$ reciprocals in AP $\Rightarrow$ middle term smaller $\Rightarrow$ use convexity for powers.
Updated On: Apr 16, 2026
- \(a^n + c^n<2b^n\)
- \(a^n + c^n>2b^n\)
- \(a^n + c^n = 2b^n\)
- None of these
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The Correct Option is B
Solution and Explanation
Concept:
If \(a, b, c\) are in HP:
\[
\frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ are in AP}
\]
\[
\Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}
\]
Step 1: Apply inequality.
Since reciprocals are in AP, original terms follow: \[ b<\frac{a+c}{2} \]
Step 2: Raise to power \(n\).
For positive \(a,b,c\): \[ b^n<\left(\frac{a+c}{2}\right)^n \] Using convexity (Jensen/AM-GM idea): \[ \left(\frac{a+c}{2}\right)^n \le \frac{a^n + c^n}{2} \] \[ \Rightarrow 2b^n<a^n + c^n \] Conclusion: \[ {a^n + c^n>2b^n} \]
Step 1: Apply inequality.
Since reciprocals are in AP, original terms follow: \[ b<\frac{a+c}{2} \]
Step 2: Raise to power \(n\).
For positive \(a,b,c\): \[ b^n<\left(\frac{a+c}{2}\right)^n \] Using convexity (Jensen/AM-GM idea): \[ \left(\frac{a+c}{2}\right)^n \le \frac{a^n + c^n}{2} \] \[ \Rightarrow 2b^n<a^n + c^n \] Conclusion: \[ {a^n + c^n>2b^n} \]
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