JEE Mains · Maths · STD 11 - 8. sequence and series
Let the positive numbers \(a _1, a _2, a _3, a _4\) and \(a _5\) be in a G.P. Let their mean and variance be \(\frac{31}{10}\) and \(\frac{ m }{ n }\) respectively, where \(m\) and \(n\) are co-prime. If the mean of their reciprocals is \(\frac{31}{40}\) and \(a_3+a_4+a_5=14\), then \(m + n\) is equal to \(.........\).
- A \(210\)
- B \(212\)
- C \(213\)
- D \(211\)
Answer & Solution
Correct Answer
(D) \(211\)
Step-by-step Solution
Detailed explanation
Let \(\frac{a}{r}, \frac{a}{r}, a, a r, a r^2\) \(\text { Given } \frac{a}{r^2}+\frac{a}{r}+a+a r+a r^2=5 \times \frac{31}{10}\) \(\text { And } \frac{r^2}{a}+\frac{r}{a}+\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^2}=5 \times \frac{31}{40}\)…
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