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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 $.........$.
$210$
$212$
$213$
$211$
Solution
Let $\frac{a}{r}, \frac{a}{r}, a, a r, a r^2$
$\text { Given } \frac{a}{r^2}+\frac{a}{r}+a+a r+a r^2=5 \times \frac{31}{10}$
$\text { And } \frac{r^2}{a}+\frac{r}{a}+\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^2}=5 \times \frac{31}{40}$
$(1) \div(2) a^2=4 \Rightarrow a=2 \quad \therefore r+\frac{1}{r}=5 / 2 \quad(a \neq-2)$
$\Rightarrow r=2$
$\therefore \text { Now } \frac{1}{2}, 1,2.4,8$
$\therefore \sigma^2=\frac{\sum x^2}{N}-\left(\frac{\sum x }{ N }\right)^2$
$=\frac{186}{25}=\frac{ M }{N} \Rightarrow 211= m + n$
