Let the positive numbers a1, a2, a3, a4, and a5 be in a G.P. Let their mean and variance be 3110 and mn, respectively, w

Question

Let the positive numbers a1, a2, a3, a4, and a5 be in a G.P. Let their mean and variance be 3110 and mn, respectively, where m and n are co-prime. If the mean of their reciprocals is 3140 and a3 + a4…

Accepted Answer

Given: The five positive numbers are in G.P. with mean $\frac{31}{10}$ , mean of reciprocals $\frac{31}{40}$ , and $a_3+a_4+a_5=14$ .

Find: The value of $m+n$ where the variance is $\frac{m}{n}$ .

Let the terms be

\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2

From the given mean,

\frac{a}{r^2}+\frac{a}{r}+a+ar+ar^2=5\times \frac{31}{10}=\frac{31}{2}

Also, from the mean of reciprocals,

\frac{r^2}{a}+\frac{r}{a}+\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2}=5\times \frac{31}{40}=\frac{31}{8}

Dividing the first equation by the second gives

a^2=4

Since all numbers are positive,

a=2

Now substituting $a=2$ into the first relation,

\frac{1}{r^2}+\frac{1}{r}+1+r+r^2=\frac{31}{4}

Hence,

\left(r+\frac{1}{r}\right)^2+\left(r+\frac{1}{r}\right)=\frac{31}{4}+1=\frac{35}{4}

Let

t=r+\frac{1}{r}

Then

t^2+t=\frac{35}{4}

so

4t^2+4t-35=0

Thus,

t=\frac{5}{2}

which gives

r=2

Therefore the numbers are

\frac{1}{2}, 1, 2, 4, 8

This also satisfies

a_3+a_4+a_5=2+4+8=14

Now,

\sigma^2=\frac{\sum x^2}{N}-\left(\frac{\sum x}{N}\right)^2

Here,

\sum x^2=\left(\frac{1}{2}\right)^2+1^2+2^2+4^2+8^2=\frac{1}{4}+1+4+16+64=\frac{341}{4}

So,

\sigma^2=\frac{\frac{341}{4}}{5}-\left(\frac{31}{10}\right)^2=\frac{341}{20}-\frac{961}{100}=\frac{1705-961}{100}=\frac{744}{100}=\frac{186}{25}

Thus,

\frac{m}{n}=\frac{186}{25}

with $m=186$ and $n=25$ . Therefore,

The required value is $211$ .

The solution image confirms the same steps:

first obtain $a^2=4$ , hence $a=2$ ,
then use $\frac{1}{r^2}+\frac{1}{r}+1+r+r^2=\frac{31}{4}$ ,
solve for $r+\frac{1}{r}=\frac{5}{2}$ and hence $r=2$ ,
get the terms $\frac{1}{2},1,2,4,8$ ,
and compute variance as $\frac{186}{25}$ .

Handwritten-style solution image showing GP terms, derivation of a equals 2, r equals 2, numbers 1 by 2, 1, 2, 4, 8, and variance simplified to 186 by 25.

Hence $m+n=186+25=211$ .

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