Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let the positive numbers and <img alt="a_5" src="https://entrancecorner.oncodec

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{\mathrm{m}}{\mathrm{n}} respectively, where \mathrm{m} \ and \ \mathrm{n} are co-prime. If the mean of their reciprocal is \frac{31}{40} and a_3+a_4+a_5=14, then \mathrm{m}+\mathrm{n}is equal to____

Option: 1

211


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

Let \frac{a}{r^2}, \frac{a}{r}, a, a r, a r^2 \\ Given \frac{a}{r^2}+\frac{a}{r}+a+a r+a r^2=5 \times \frac{31}{10} \ldots (1)\\ 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} \ldots (2)
\begin{aligned} & (1) \div(2) a^2=4 \Rightarrow a=2 \\ & \therefore r+\frac{1}{r}=5 / 2(a \neq-2) \\ & \Rightarrow r=2 \\ & \therefore \text { Now } \frac{1}{2}, 1,2,4,8 \end{aligned}

\begin{aligned} & \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 \end{aligned}

 

Posted by

Ramraj Saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile
anam-khan

Low JEE Main rank? Explore alternative paths for engineering admission
anam-khan

Gender-gap in JEE Main 2025: Female participation remains around 31%, overall pool drops by 7.09%

Latest Question