Question

Stats help required

Answer 1

Any hints?

Answer 2

sorry, don't even know what GM is

Answer 3

geometric mean?

Answer 4

Yeah lol :P

Answer 5

Do you know the formula for GM @hba ??

Answer 6

The GM of \(a,b,c\cdots \) is \(\sqrt{abc\cdots}\)

Answer 7

So G1.G2?

Answer 8

I also got the same thing but i don't think that would be the answer to my problem.

Answer 9

It seems like that is the answer.

Answer 10

You sure?

Answer 11

I don't think so.Moreover you have made a few mistakes above.

Answer 12

\[\sqrt[n].\]

Answer 13

Well first of all that would not be a square root.It would be G1=(a1.a2.....an)^(1/n)

Answer 14

Yeah

Answer 15

Frook.

Answer 16

Also,

G2=(b1.b2........bm)^(1/m)

Answer 17

@mathslover

Yes i do.

Answer 18

Is your doubt clear now hba?

Answer 19

I don't think so i have any doubts.I just don't know how to do this freaking question -_-

Answer 20

Well i don't think so the solution is that simple.

Answer 21

Yes, I am tryig that.

Answer 22

I am waiting.

Answer 23

I am getting :

\(\large{G_{m+n} = (\cfrac{G_n^m}{G_m^n})^{\cfrac{1}{m-n}}}\)

Answer 24

Hba , do you have the answer?

Answer 25

No

Answer 26

@mathslover : how did you get the answer?

Answer 27

5 minute please

Answer 28

if
\[ G_{1} = \sqrt[n]{a_{1} . a_{2} ... a_{n} } \]
and \(G_{2} = \sqrt[m]{b_{1} . b_{2} ... b_{m} }\)

aren't we looking for
\[\sqrt[n+m]{a_{1} . a_{2} ... a_{n} .b_{1} . b_{2} ... b_{m} }\] ?

Answer 29

Let a G.P : a, ar , ...

\(\large{G_m = [a * ar * ar^2 ... * ar^{m-1}]^\cfrac{1}{m}} \)

\(G_m = ( a^ m . r^{\cfrac{(m-1)m}{2}}) ^\cfrac{1}{m}\)

Answer 30

\(G_ m = a . r^(\cfrac{m-1}{2})\)

\(\implies a = \cfrac{G_m}{r^\cfrac{m-1}{2}}\)

\(\large G_n = a. r^ \cfrac{n-1}{2}\)

\(\large G_n = G_m . r^{(\cfrac{n-1}{2} - \cfrac{m-1}{2})}\)

Answer 31

\(G_n = G_m \times r ^{(\cfrac{n-m}{2})}\)

\(G_{m+n} = a . r^{(\cfrac{m+n-1}{2})}\)

\(\implies G_m . r^{(\cfrac{m+n-1}{2} - \cfrac{m-1}{2})}\)

\(\implies G_m \times r^{\cfrac{n}{2}}\)

\(\implies G_m \times {(\cfrac{G_n}{G_m})}^{\cfrac{n}{n-m}}\)

\(\implies G_n ^{\cfrac{n}{n-m}} \times G_m ^ {\cfrac{-m}{n-m}}\)

\(\implies [\cfrac{G_m^m}{G_n^n}]^{(\cfrac{1}{m-n})}\)

\(\implies [\cfrac{G_n^n}{G_m^m}] ^ \cfrac{1}{n-m}\)

Answer 32

This is what I did friends, lemme know if it helps any one :)

Answer 33

@chihiroasleaf ? Any views?

Answer 34

@Hero

Do you think it's correct?

Answer 35

@NadiaKamal

Answer 36

in which step are you getting problem Hba?

Answer 37

@hartnn: Mind helping.

Answer 38

@hartnn
geometric mean of m numbers is Mth root of x1 to xm

Answer 39

Its not sqrt

Answer 40

https://en.wikipedia.org/wiki/Geometric_mean

Have a look

Answer 41

Moreover, the last expression is wrong. (I tried it using 2 examples)

Answer 42

I understand :)

Answer 43

\(\large GM_m = \sqrt[m]{x1.x2......xm} \implies GM_m^m=x1.x2......xm\)
\(\large GM_n = \sqrt[n]{y1.y2......ym}\implies GM_n^n=y1.y2......ym\)

\(\large GM_{m+n} = \sqrt[m+n]{(x1.x2......xm)(y1.y2......ym)} \\ \large = \sqrt[m+n]{GM_m^m .GM_n^n} \\ \Huge = (GM_m)^{\dfrac{m}{m+n}}(GM_n)^{\dfrac{n}{m+n}}\)

@hba ask if any doubts

Answer 44

@hartnn Thanks a lot hartnn :D