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WBJEE · Maths · Sequences and Series

Let \(a_1, a_2, a_3, \ldots, a_n\) be positive real numbers. Then the minimum value of \(\frac{a_1}{a_2}+\frac{a_2}{a_3}+\ldots .+\frac{a_n}{a_1}\) is

  1. A 1
  2. B \(\mathrm{n}\)
  3. C \({ }^n \mathrm{C}_2\)
  4. D 2
Verified Solution

Answer & Solution

Correct Answer

(B) \(\mathrm{n}\)

Step-by-step Solution

Detailed explanation

Hint: \(\frac{\frac{a_1}{a_2}+\frac{a_2}{a_3}+\ldots . .+\frac{a_n}{a_1}}{n} \geq \sqrt[n]{\frac{a_1}{a_2} \times \frac{a_2}{a_3} \times \ldots . \frac{a_n}{a_1}}\) \(A M \geq G M\) (property) \(\therefore\) Minimum value \(=\mathrm{n}\) It is possible when…