Question

Help please!

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​

Answer 1

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

Answer 2

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}`