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Prove that,
when the values in a database are equal to each other, then the A.M, G.M and H.M equal to each other
note:
A.M=arithmetic mean
G.M=geometric mean
H.M= harmonic mean​

User Jacob K
by
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2 Answers

4 votes

Answer:

What
\colorbox{red}{Nayefx}says is I say

User Abhishek Tomar
by
8.2k points
4 votes

Answer:

See below

Explanation:

the n number of value of x


\displaystyle x_(1),x _(2) \dots x_(n)

let it be


\displaystyle x_(1) = x _(2) = x_(3){\dots }= x_(n) = a

now, the A.M of x is


\rm \displaystyle \: A.M = ( x_(1) + x_(2) + \dots \dots \: + x_(n) )/(n)

since every value equal to a

substitute:


\rm \displaystyle \: A.M = ( a + a + \dots \dots \: + a)/(n)


\rm \displaystyle \: A.M = ( na)/(n)

reduce fraction:


\rm \displaystyle \: A.M = a

the G.M of x is


\rm\displaystyle \: G.M =( x_(1) * x _(2) {\dots }* x_(n) {)}^{ {1}^{}/ {n}^{} }

since every value equal to a

substitute:


\rm\displaystyle \: G.M =( a * a{\dots }* a{)}^{ {1}^{}/ {n}^{} }

recall law of exponent:


\rm\displaystyle \: G.M =( {a}^(n) {)}^{ {1}^{}/ {n}^{} }

recall law of exponent:


\rm\displaystyle \: G.M = a

the H.M of x is


\displaystyle \: H.M = \frac{n}{ (1)/( x_(1)) + (1)/( x_(2) ) {\dots } \: { \dots}(1)/(x _(n) ) }

since every value equal to a

substitute:


\displaystyle \: H.M = \frac{n}{ (1)/( a) + (1)/( a ) {\dots } \: { \dots}(1)/(a ) }


\displaystyle \: H.M = (n)/( (n)/(a) )

simplify complex fraction:


\displaystyle \: H.M = n * (a)/(n)


\displaystyle \: H.M = a \:

so


\displaystyle \: A.M = G.M = H.M = a

hence,


\text{Proven}

User Dineshthamburu
by
8.0k points

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