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Statistics question

2023 · 12 Apr · Shift 1 · Q42
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Statistics question

2023 · 12 Apr · Shift 1 · Q42

JEE MainMathematicsStatisticsNumerical+4 / −1
Let the positive numbers a1,a2,a3,a4a_{1}, a_{2}, a_{3}, a_{4}a1​,a2​,a3​,a4​ and a5a_{5}a5​ be in a G.P. Let their mean and variance be 3110\frac{31}{10}1031​ and mn\frac{m}{n}nm​ respectively, where mmm and nnn are co-prime. If the mean of their reciprocals is 3140\frac{31}{40}4031​ and a3+a4+a5=14a_{3}+a_{4}+a_{5}=14a3​+a4​+a5​=14, then m+nm+nm+n is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 211

  1. Set up the G.P.

Let the five positive numbers in G.P. be

xr2,  xr,  x,  xr,  xr2\frac{x}{r^2},\; \frac{x}{r},\; x,\; xr,\; xr^2r2x​,rx​,x,xr,xr2

with x>0x>0x>0 and r>0r>0r>0.

  1. Use the mean of the numbers

Given arithmetic mean is 3110\dfrac{31}{10}1031​, so

15(xr2+xr+x+xr+xr2)=3110\frac{1}{5}\left(\frac{x}{r^2}+\frac{x}{r}+x+xr+xr^2\right)=\frac{31}{10}51​(r2x​+rx​+x+xr+xr2)=1031​

which gives

x(r2+r+1+1r+1r2)=312.(1)x\left(r^2+r+1+\frac1r+\frac1{r^2}\right)=\frac{31}{2}. \tag{1}x(r2+r+1+r1​+r21​)=231​.(1)
  1. Use the mean of reciprocals

The reciprocals are also in G.P.:

r2x,  rx,  1x,  1xr,  1xr2\frac{r^2}{x},\; \frac{r}{x},\; \frac1x,\; \frac1{xr},\; \frac1{xr^2}xr2​,xr​,x1​,xr1​,xr21​

Their mean is 3140\dfrac{31}{40}4031​, so

15(r2x+rx+1x+1xr+1xr2)=3140\frac{1}{5}\left(\frac{r^2}{x}+\frac{r}{x}+\frac1x+\frac1{xr}+\frac1{xr^2}\right)=\frac{31}{40}51​(xr2​+xr​+x1​+xr1​+xr21​)=4031​

Hence

1x(r2+r+1+1r+1r2)=318.(2)\frac{1}{x}\left(r^2+r+1+\frac1r+\frac1{r^2}\right)=\frac{31}{8}. \tag{2}x1​(r2+r+1+r1​+r21​)=831​.(2)

Let

S=r2+r+1+1r+1r2.S=r^2+r+1+\frac1r+\frac1{r^2}.S=r2+r+1+r1​+r21​.

Then from (1) and (2):

xS=312,Sx=318.xS=\frac{31}{2}, \qquad \frac{S}{x}=\frac{31}{8}.xS=231​,xS​=831​.

Multiplying,

S2=312⋅318=96116S^2=\frac{31}{2}\cdot \frac{31}{8}=\frac{961}{16}S2=231​⋅831​=16961​

so, since S>0S>0S>0,

S=314.S=\frac{31}{4}.S=431​.

Then from xS=312xS=\frac{31}{2}xS=231​,

x⋅314=312  ⟹  x=2.x\cdot \frac{31}{4}=\frac{31}{2}\implies x=2.x⋅431​=231​⟹x=2.

So the terms are

2r2,  2r,  2,  2r,  2r2.\frac{2}{r^2},\; \frac{2}{r},\; 2,\; 2r,\; 2r^2.r22​,r2​,2,2r,2r2.
  1. Use a3+a4+a5=14a_3+a_4+a_5=14a3​+a4​+a5​=14

Given

a3+a4+a5=2+2r+2r2=14.a_3+a_4+a_5=2+2r+2r^2=14.a3​+a4​+a5​=2+2r+2r2=14.

Thus

1+r+r2=71+r+r^2=71+r+r2=7

so

r2+r−6=0r^2+r-6=0r2+r−6=0

which gives

(r−2)(r+3)=0.(r-2)(r+3)=0.(r−2)(r+3)=0.

Since r>0r>0r>0, we get

r=2.r=2.r=2.

Therefore the numbers are

24,  22,  2,  4,  8\frac{2}{4},\; \frac{2}{2},\; 2,\; 4,\; 842​,22​,2,4,8

that is,

12,  1,  2,  4,  8.\frac12,\; 1,\; 2,\; 4,\; 8.21​,1,2,4,8.
  1. Compute the variance

Mean is

μ=3110.\mu=\frac{31}{10}.μ=1031​.

Variance is

σ2=15∑ai2−μ2.\sigma^2=\frac{1}{5}\sum a_i^2-\mu^2.σ2=51​∑ai2​−μ2.

Now

a12+a22+a32+a42+a52=(12)2+12+22+42+82=14+1+4+16+64=3414.a_1^2+a_2^2+a_3^2+a_4^2+a_5^2 =\left(\frac12\right)^2+1^2+2^2+4^2+8^2 =\frac14+1+4+16+64 =\frac{341}{4}.a12​+a22​+a32​+a42​+a52​=(21​)2+12+22+42+82=41​+1+4+16+64=4341​.

Hence

15∑ai2=34120.\frac{1}{5}\sum a_i^2=\frac{341}{20}.51​∑ai2​=20341​.

Also

μ2=(3110)2=961100.\mu^2=\left(\frac{31}{10}\right)^2=\frac{961}{100}.μ2=(1031​)2=100961​.

Therefore

σ2=34120−961100=1705−961100=744100=18625.\sigma^2=\frac{341}{20}-\frac{961}{100} =\frac{1705-961}{100} =\frac{744}{100} =\frac{186}{25}.σ2=20341​−100961​=1001705−961​=100744​=25186​.

So

mn=18625\frac{m}{n}=\frac{186}{25}nm​=25186​

with m=186m=186m=186, n=25n=25n=25.

Thus

m+n=186+25=211.m+n=186+25=211.m+n=186+25=211.
  1. Compare with stored answer

Derived answer is 211211211, which matches the stored correct answer.

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