If M is the mean of x₁, x₂, x₃, x₄, x₅ and x₆, prove that
(x₁ − M) + (x₂ − M) (x₃ − M) (x₄ − M) + (x₅ − M) + (x₆ − M) = 0.

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Solution

Given that the mean of x₁, x₂, x₃, x₄, x₅, and x₆ is M.

We have to prove that

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore, we have

Now,

Hence the proof is complete.

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Similar questions

$I f M i s t h e m e a n o f x_{1}, x_{2}, x_{3}, x_{4}, x_{5} a n d x_{6}, p r o v e t h a t (x_{1} - M) + (x_{2} - M) + (x_{3} - M) + (x_{4} - M) + (x_{5} - M) + (x_{6} - M) = 0$

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α

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is-

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\frac{x}{6}

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If M is the mean of x1, x2, x3, x4, x5 and x6, prove that (x1 − M) + (x2 − M) (x3 − M) (x4 − M) + (x5 − M) + (x6 − M) = 0.

Given that the mean of x1, x2, x3, x4, x5, and x6 is M. We have to prove that Remember the definition of mean of n values x1, x2… xn is Therefore, we have Now, Hence the proof is complete.

If M is the mean of x₁, x₂, x₃, x₄, x₅ and x₆, prove that
(x₁ − M) + (x₂ − M) (x₃ − M) (x₄ − M) + (x₅ − M) + (x₆ − M) = 0.

Given that the mean of x₁, x₂, x₃, x₄, x₅, and x₆ is M.

We have to prove that

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore, we have

Now,

Hence the proof is complete.