Let the positive numbers a_1,a_2,… - JEE Main 2023 Mathematics PYQ

JEE Main 2023 — Mathematics Probability & Statistics

2023integerhard

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 $\frac{31}{10}$ and ${a}_{3}+{a}_{4}+{a}_{5}=14$ , then $m+n$ is equal to ____________.

Official previous-year question

Held on 12 Apr 2023 · Verified 6 Jul 2026.

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Solution

$5$ positive numbers ${a}_{1},{a}_{2},...,{a}_{5}$ are in geometric progression. So, let the numbers be $\frac{a}{{r}^{2}},\frac{a}{r},a,ar,a{r}^{2}$

According to question,

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

And,

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

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

By $(1)&(2)$ , we get

${a}^{2}=4\Rightarrow a=2$

So,

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

$\Rightarrow {(r+\frac{1}{r})}^{2}+r+\frac{1}{r}=\frac{27}{4}+2$

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

where, $t=r+\frac{1}{r}$

$4{t}^{2}+4t-35=0$

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

So, numbers are $\frac{1}{2},1,2,4,8$

Variance is

$=\frac{\frac{1}{4}+1+4+16+64}{5}-{(\frac{\frac{1}{2}+1+2+4+8}{5})}^{2}$

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

$m+n=211$

Hence this is the required option.

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