Let the positive numbers a_1, a_2, a_3, a_4 a | vedclass

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(D) Let the terms be $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$. Given the mean is $\frac{31}{10}$,so $\frac{a}{r^2} + \frac{a}{r} + a + ar + ar^2 = 5

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$211$

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(D) Let the terms be $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$. Given the mean is $\frac{31}{10}$,so $\frac{a}{r^2} + \frac{a}{r} + a + ar + ar^2 = 5 	imes \frac{31}{10} = \frac{31}{2}$. Given the mean of reciprocals is $\frac{31}{40}$,so $\frac{r^2}{a} + \frac{r}{a} + \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} = 5 	imes \frac{31}{40} = \frac{31}{8}$. Dividing the first equation by the second,we get $a^2 = \frac{31/2}{31/8} = 4$,so $a = 2$ (since terms are positive). Substituting $a=2$ into the first equation: $\frac{2}{r^2} + \frac{2}{r} + 2 + 2r + 2r^2 = \frac{31}{2}$ $\Rightarrow r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2} = \frac{31}{4}$. Let $x = r + \frac{1}{r}$. Then $x^2 - 2 + x + 1 = \frac{31}{4} \Rightarrow x^2 + x - \frac{27}{4} = 0$. Solving for $x$,we find $x = \frac{5}{2}$,so $r + \frac{1}{r} = \frac{5}{2} \Rightarrow r = 2$ or $r = 1/2$. Given $a_3+a_4+a_5 = a+ar+ar^2 = 2+2r+2r^2 = 14$ $\Rightarrow r^2+r-6=0$ $\Rightarrow (r+3)(r-2)=0$. Since $r>0$,$r=2$. The terms are $\frac{1}{2}, 1, 2, 4, 8$. Mean $\bar{x} = \frac{31}{10}$. Variance $\sigma^2 = \frac{\sum a_i^2}{5} - (\bar{x})^2 = \frac{1/4 + 1 + 4 + 16 + 64}{5} - (\frac{31}{10})^2 = \frac{85.25}{5} - \frac{961}{100} = 17.05 - 9.61 = 7.44 = \frac{744}{100} = \frac{186}{25}$. Thus $m=186, n=25$,and $m+n = 211$.

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 $.........$.

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