The sum of the squares of three consecutive positive numbers is 245. Find the numbers.

Let the three consecutive positive numbers be \(x - 1\), \(x\), and \(x + 1\).

The sum of the squares of these three numbers is given as 245. So, we can write the equation:

\[
(x - 1)^2 + x^2 + (x + 1)^2 = 245
\]

Now, expand and simplify each square:

\[
(x - 1)^2 = x^2 - 2x + 1
\]
\[
x^2 = x^2
\]
\[
(x + 1)^2 = x^2 + 2x + 1
\]

Adding these up, the equation becomes:

\[
(x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) = 245
\]

Combine like terms:

\[
x^2 - 2x + 1 + x^2 + x^2 + 2x + 1 = 245
\]

\[
3x^2 + 2 = 245
\]

Subtract 2 from both sides:

\[
3x^2 = 243
\]

Divide by 3:

\[
x^2 = 81
\]

Take the square root of both sides:

\[
x = 9
\]

Since \(x = 9\), the three consecutive positive numbers are:

\[
x - 1 = 8, \quad x = 9, \quad x + 1 = 10
\]

Thus, the three consecutive positive numbers are **8, 9, and 10**.

### Verification

Let's verify by calculating the sum of their squares:

\[
8^2 + 9^2 + 10^2 = 64 + 81 + 100 = 245
\]

The calculation is correct. Therefore, the numbers are indeed **8, 9, and 10**.

Would you like more details or have any questions?

Here are some related questions you might find interesting:
1. What is the sum of squares of four consecutive numbers that adds up to 354?
2. How can we generalize the formula for the sum of squares of \(n\) consecutive numbers?
3. Can the sum of cubes of three consecutive numbers be a perfect square?
4. How would you solve for three consecutive numbers if given their product instead of the sum of squares?
5. What other patterns emerge with consecutive numbers when summing their squares or cubes?

**Tip:** Always verify your solution by plugging the numbers back into the original equation to ensure they satisfy the given conditions.

Learn how to find three consecutive positive numbers whose sum of squares equals 245. Follow a step-by-step solution with detailed calculations and verification. Suitable for Grades 7-9.

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