If ` a + b + c = m and (1)/( a) + (1)/(b) + (1)/( c) = O` then average of ` a^(2) , b^(2) and c^(2)` is

Options

A. `m^(2)`
B. `(m^(2))/(3)`
C. `(m^(2))/(9)`
D. `(m^(2))/(27)`

Answer

To solve the problem, we need to find the average of \( a^2, b^2, \) and \( c^2 \) given the equations:

1. \( a + b + c = m \)
2. \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \)

### Step-by-Step Solution:

**Step 1: Square the first equation.**

We start with the equation \( a + b + c = m \). To find \( a^2 + b^2 + c^2 \), we square both sides:

\[ (a + b + c)^2 = m^2 \]

**Hint:** Remember that squaring a sum expands to include the squares of each term and twice the product of each pair.

**Step 2: Expand the squared equation.**

Expanding the left side, we have:

\[ a^2 + b^2 + c^2 + 2(ab + bc + ca) = m^2 \]

**Hint:** Use the formula \( (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) \) for expansion.

**Step 3: Isolate \( a^2 + b^2 + c^2 \).**

We want to isolate \( a^2 + b^2 + c^2 \):

\[ a^2 + b^2 + c^2 = m^2 - 2(ab + bc + ca) \]

**Hint:** Rearranging equations is key to isolating the variable of interest.

**Step 4: Analyze the second equation.**

From the second equation \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \), we can rewrite it as:

\[ \frac{bc + ca + ab}{abc} = 0 \]

This implies that:

\[ bc + ca + ab = 0 \]

**Hint:** When the sum of fractions equals zero, the numerator must be zero.

**Step 5: Substitute back into the equation.**

Now, substituting \( ab + bc + ca = 0 \) into our equation for \( a^2 + b^2 + c^2 \):

\[ a^2 + b^2 + c^2 = m^2 - 2(0) = m^2 \]

**Hint:** Substituting known values simplifies the equation.

**Step 6: Calculate the average.**

To find the average of \( a^2, b^2, \) and \( c^2 \):

\[ \text{Average} = \frac{a^2 + b^2 + c^2}{3} = \frac{m^2}{3} \]

**Hint:** The average is simply the sum divided by the number of terms.

### Final Answer:

The average of \( a^2, b^2, \) and \( c^2 \) is:

\[ \frac{m^2}{3} \]