If average of a,b,c is m and (ab+bc+ca)=0, then the average of a2,b2,c2 is :m23m29m2m
Question
If average of a, b, c is m and (ab + bc + ca) = 0, then the average of a², b², c² is :
m
23
m
29
m²
m
Solution
The question seems to be incorrectly formatted, making it difficult to understand what is being asked. However, I'll try to interpret it as best as I can.
Given that the average of a, b, c is m, we can write this as:
(a + b + c) / 3 = m => a + b + c = 3m -----(1)
We are also given that ab + bc + ca = 0.
We need to find the average of a^2, b^2, c^2. Let's denote this as n for simplicity, so:
(a^2 + b^2 + c^2) / 3 = n => a^2 + b^2 + c^2 = 3n -----(2)
We can express a^2 + b^2 + c^2 as (a + b + c)^2 - 2(ab + bc + ca) using the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).
Substituting the given values from equation (1) and the given condition ab + bc + ca = 0, we get:
a^2 + b^2 + c^2 = (3m)^2 - 2*0 => a^2 + b^2 + c^2 = 9m^2
So, the average of a^2, b^2, c^2 is 9m^2 / 3 = 3m^2.
Therefore, the answer is 3m^2.
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