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 31 10 and m n respectively, where m and n are co-prime. If the mean of their reciprocals is 31 40 and a 3 + a 4 + a 5 = 14 , then m + n is

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Solution

Q.
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 _______ . [2023]

Ans.
(211)

$Let \frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2} be positive numbers in G.P.$

$\Rightarrow \frac{a}{r^{2}} + \frac{a}{r} + a + a r + a r^{2} = 5 \times \frac{31}{10}$ ...(i)

$and$ $\frac{r^{2}}{a} + \frac{r}{a} + \frac{1}{a} + \frac{1}{a r} + \frac{1}{a r^{2}} = 5 \times \frac{31}{40}$ ...(ii)

$Divide (i) by (ii), we get$

$\frac{a (\frac{1}{r^{2}} + \frac{1}{r} + 1 + r + r^{2})}{\frac{1}{a} (r^{2} + r + 1 + \frac{1}{r} + \frac{1}{r^{2}})} = \frac{5 \times \frac{31}{10}}{5 \times \frac{31}{40}}$

$\Rightarrow a^{2} = 4$

$\Rightarrow a = 2$ $(∵ a cannot be negative)$

$From (i), a (\frac{1}{r^{2}} + \frac{1}{r} + 1 + r + r^{2}) = 5 \times \frac{31}{10}$

$\Rightarrow {(r + \frac{1}{r})}^{2} + (r + \frac{1}{r}) = \frac{31}{4} + 1$

$\Rightarrow 4 t^{2} + 4 t - 35 = 0$ $[Let, r + \frac{1}{r} = t]$

$\Rightarrow t = \frac{5}{2} \Rightarrow r = 2$

$∴$ $Numbers are = \frac{1}{2}, 1, 2, 4, 8$

$∴$ $σ^{2} = \frac{\sum x^{2}}{N} - {(\frac{\sum x}{N})}^{2} = \frac{\frac{1}{4} + 1 + 4 + 16 + 64}{5} - {(\frac{31}{10})}^{2}$

$= \frac{341}{20} - \frac{961}{100} = \frac{186}{25}$ $∴$ $m + n = 186 + 25 = 211$

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