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QJEEM23Mace05: JEE Main 2023 Mathematics - Class 11 JEEM23Mace05 Solution: Let the positive numbers and be in a G.P. Let their mean and…

Let the positive numbers a1,a2,a3,a4a_{1}, a_{2}, a_{3}, a_{4} and a5a_{5} be in a G.P. Let their mean and variance be 3110\frac{31}{10} and mn\frac{m}{n} respectively, where mm and nn are co-prime. If the mean of their reciprocals is 3140\frac{31}{40} and a3+a4+a5=14a_{3}+a_{4}+a_{5}=14 , then m+nm+n is equal to ___________.

Step-by-Step Verified Solution

Since a1,a2,a3,a4,a5a_1, a_2, a_3, a_4, a_5 are in geometric progression, we can write :

a2=raa_2 = r a ,

a3=r2aa_3 = r^2 a ,

a4=r3aa_4 = r^3 a ,

a5=r4aa_5 = r^4 a .

where rr is the common ratio and a1a_1 = aa is the first term.

Given that the mean of the series is 3110\frac{31}{10} , we get

15(a1+a2+a3+a4+a5)=3110\frac{1}{5}(a_1 + a_2 + a_3 + a_4 + a_5) = \frac{31}{10}

Substituting the terms with the values of a1a_1 and rr gives

15(a+ra+r2a+r3a+r4a)=3110\frac{1}{5}(a + r a + r^2 a + r^3 a + r^4 a) = \frac{31}{10}

Simplifying this gives

a(1+r+r2+r3+r4)=312a (1 + r + r^2 + r^3 + r^4) = \frac{31}{2}

amp;a(r5−1)r−1=312......(1)amp;1a(1+1r+1r2+1r3+1r4)=3140⋅5=318amp;1a(1−(1r)51−1r)=318amp; or 1a(r5−1r−1)1r4=318.........(2)\begin{aligned}& \frac{a\left(r^5-1\right)}{r-1}=\frac{31}{2} ......(1) \\\\& \frac{1}{a}\left(1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+\frac{1}{r^4}\right)=\frac{31}{40} \cdot 5=\frac{31}{8} \\\\& \frac{1}{a}\left(\frac{1-\left(\frac{1}{r}\right)^5}{1-\frac{1}{r}}\right)=\frac{31}{8} \\\\& \text { or } \frac{1}{a}\left(\frac{r^5-1}{r-1}\right) \frac{1}{r^4}=\frac{31}{8} .........(2)\end{aligned}

From (1) and (2)

1a⋅312a⋅1r4=318\frac{1}{a} \cdot \frac{31}{2 a} \cdot \frac{1}{r^4}=\frac{31}{8}

ar2=2a r^2=2

From (1)

amp;2r2(r5−1r−1)=312amp;1+r+r2+r3+r4r2=314amp;(r2+1r2)+(r+1r)=274amp;t2−2+t=274\begin{aligned}& \frac{2}{r^2}\left(\frac{r^5-1}{r-1}\right)=\frac{31}{2} \\\\& \frac{1+r+r^2+r^3+r^4}{r^2}=\frac{31}{4} \\\\& \left(r^2+\frac{1}{r^2}\right)+\left(r+\frac{1}{r}\right)=\frac{27}{4} \\\\& t^2-2+t=\frac{27}{4}\end{aligned}

amp;4t2+4t−35=0amp;4t2+14t−10t−35=0amp;(2t−5)(2t+7)=0amp;t=52,−72⇒r=2amp;∴r=2,a=12\begin{aligned}&4 t^2+4 t-35=0\\\\&4 t^2+14 t-10 t-35=0\\\\&(2 t-5)(2 t+7)=0\\\\&t=\frac{5}{2}, \frac{-7}{2} \Rightarrow r=2\\\\& \therefore r=2, a=\frac{1}{2}\end{aligned}

 Variance of data set {12,1,2,4,8}\text { Variance of data set }\left\{\frac{1}{2}, 1,2,4,8\right\}

∴σ2=∑X2 N−(∑XN)2\therefore \sigma^2=\frac{\sum{\mathrm{X}^2}}{\mathrm{~N}}-\left(\frac{\sum{\mathrm{X}}}{\mathrm{N}}\right)^2

amp;=(3414)5−(3110)2amp;=34120−961100=1705−961100amp;=744100\begin{aligned}& =\frac{\left(\frac{341}{4}\right)}{5}-\left(\frac{31}{10}\right)^2 \\\\& =\frac{341}{20}-\frac{961}{100}=\frac{1705-961}{100} \\\\& =\frac{744}{100}\end{aligned}

=18625=mn⇒211=m+n=\frac{186}{25}=\frac{\mathrm{m}}{\mathrm{n}} \Rightarrow 211=\mathrm{m}+\mathrm{n}

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