If the arithmetic mean of a, b, c is \(\rm \frac M 3\) and \(\rm \frac{1}{a} + \frac{1}{b} = -\frac{1}{c} \) , then the arithmetic mean of a 2, b 2, c 2 is
M 2/3
The arithmetic mean (or average) of a set of numbers is found by summing the numbers and dividing by the count of numbers in the set.
We are given two pieces of information:
a, b, and c is \(\rm \frac{M}{3}\).\(\rm \frac{1}{a} + \frac{1}{b} = -\frac{1}{c}\).From the definition of arithmetic mean:
\(\rm \frac{a + b + c}{3} = \frac{M}{3}\)
Multiplying both sides by 3 gives:
\(\rm a + b + c = M\)
Start with the second given condition:
\(\rm \frac{1}{a} + \frac{1}{b} = -\frac{1}{c}\)
Combine the terms on the left side using a common denominator (ab):
\(\rm \frac{b + a}{ab} = -\frac{1}{c}\)
Cross-multiply:
\(\rm c(a + b) = -ab\)
Expand the left side:
\(\rm ac + bc = -ab\)
Rearrange the terms to one side:
\(\rm ab + bc + ac = 0\)
We need to find the arithmetic mean of a2, b2, and c2, which is \(\rm \frac{a^2 + b^2 + c^2}{3}\).
We know the identity:
\(\rm (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac)\)
Substitute the known values into this identity:
a + b + c = Mab + bc + ac = 0So, the equation becomes:
\(\rm (M)^2 = a^2 + b^2 + c^2 + 2(0)\)
\(\rm M^2 = a^2 + b^2 + c^2\)
Now substitute this result back into the formula for the arithmetic mean of the squares:
Arithmetic Mean = \(\rm \frac{a^2 + b^2 + c^2}{3}\)
Arithmetic Mean = \(\rm \frac{M^2}{3}\)
The arithmetic mean of a2, b2, and c2 is \(\rm \frac{M^2}{3}\).
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