JEE Main 2023 (Online) 12th April Morning ShiftQuestion 1
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 $\_\_\_\_\_\_$ .
Sequence and Series · Geometric Progression
Answer$211$
Solution
Formulas used. (1) In a $5$-term G.P., $a_1a_5=a_2a_4=a_3^2$, so $\sum\frac1{a_i}=\frac{\sum a_i}{a_3^2}$(2) Variance $=\frac{\sum a_i^2}{N}-(\text{mean})^2$Step 1 - Why formula (1) holds. $\dfrac1{a_1}=\dfrac{a_5}{a_1a_5}=\dfrac{a_5}{a_3^2}$, and likewise each $\dfrac1{a_i}=\dfrac{a_{6-i}}{a_3^2}$. Step 2 - Middle term. $\sum a_i=5\cdot\dfrac{31}{10}=\dfrac{31}{2}$ and $\sum\dfrac1{a_i}=5\cdot\dfrac{31}{40}=\dfrac{31}{8}$: $$a_3^2=\frac{31/2}{31/8}=4\ \Longrightarrow\ a_3=2 .$$ Step 3 - Ratio. $$a_3+a_4+a_5=2\left(1+r+r^2\right)=14\ \Longrightarrow\ r^2+r-6=0\ \Longrightarrow\ r=2 .$$ The numbers are $\tfrac12,1,2,4,8$ (sum $15.5$, as required). Step 4 - Variance. $$\frac{\sum a_i^2}{5}=\frac{\tfrac14+1+4+16+64}{5}=\frac{85.25}{5}=17.05,\qquad \sigma^2=17.05-(3.1)^2=7.44=\frac{186}{25}.$$ Step 5. $\gcd(186,25)=1$: $$m+n=186+25=211 .$$ Answer: 211