The sum of the squares of 3 consecutive positive numbers is 365. The sum of the numbers is

Options

A. 30
B. 33
C. 36
D. 45

Answer

To solve the problem, we need to find three consecutive positive numbers whose squares sum up to 365. Let's denote the middle number of these three consecutive numbers as \( x \). Therefore, the three consecutive numbers can be represented as:

- First number: \( x - 1 \)
- Middle number: \( x \)
- Last number: \( x + 1 \)

### Step 1: Set up the equation
The sum of the squares of these three numbers can be expressed as:
\[ (x - 1)^2 + x^2 + (x + 1)^2 = 365 \]

### Step 2: Expand the squares
Now, let's expand each term:
\[ (x - 1)^2 = x^2 - 2x + 1 \]
\[ x^2 = x^2 \]
\[ (x + 1)^2 = x^2 + 2x + 1 \]

### Step 3: Combine the expanded terms
Now, we can combine these expanded terms:
\[ (x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) = 365 \]
This simplifies to:
\[ x^2 - 2x + 1 + x^2 + x^2 + 2x + 1 = 365 \]
Combining like terms gives:
\[ 3x^2 + 2 = 365 \]

### Step 4: Isolate the variable
Next, we isolate \( 3x^2 \):
\[ 3x^2 = 365 - 2 \]
\[ 3x^2 = 363 \]

### Step 5: Solve for \( x^2 \)
Now, divide both sides by 3:
\[ x^2 = \frac{363}{3} = 121 \]

### Step 6: Find \( x \)
Taking the square root of both sides, we find:
\[ x = \sqrt{121} = 11 \]

### Step 7: Identify the three consecutive numbers
Now that we have \( x \), we can find the three consecutive numbers:
- First number: \( x - 1 = 11 - 1 = 10 \)
- Middle number: \( x = 11 \)
- Last number: \( x + 1 = 11 + 1 = 12 \)

### Step 8: Calculate the sum of the numbers
Finally, we calculate the sum of these three numbers:
\[ 10 + 11 + 12 = 33 \]

Thus, the sum of the three consecutive positive numbers is **33**.

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