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
- A 1
- B \(\mathrm{n}\)
- C \({ }^n \mathrm{C}_2\)
- D 2
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…
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