If the square of sum of three positive consecutive natural numbers exceeds the sum of their squares by 292, then what is the largest of the three numbers ?

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the problem step by step, let's denote the three consecutive natural numbers as \( n-1, n, n+1 \).

### Step 1: Define the numbers
Let the three consecutive natural numbers be:
- First number: \( n - 1 \)
- Second number: \( n \)
- Third number: \( n + 1 \)

### Step 2: Calculate the sum of the numbers
The sum of these three numbers is:
\[ (n - 1) + n + (n + 1) = 3n \]

### Step 3: Calculate the square of the sum
The square of the sum is:
\[ (3n)^2 = 9n^2 \]

### Step 4: Calculate the sum of the squares of the numbers
The sum of the squares of these three numbers is:
\[ (n - 1)^2 + n^2 + (n + 1)^2 \]
Calculating each square:
- \( (n - 1)^2 = n^2 - 2n + 1 \)
- \( n^2 = n^2 \)
- \( (n + 1)^2 = n^2 + 2n + 1 \)

Adding these together:
\[ (n^2 - 2n + 1) + n^2 + (n^2 + 2n + 1) = 3n^2 + 2 \]

### Step 5: Set up the equation
According to the problem, the square of the sum exceeds the sum of the squares by 292:
\[ 9n^2 = (3n^2 + 2) + 292 \]

### Step 6: Simplify the equation
Rearranging gives:
\[ 9n^2 = 3n^2 + 294 \]
Subtract \( 3n^2 \) from both sides:
\[ 6n^2 = 294 \]

### Step 7: Solve for \( n^2 \)
Dividing both sides by 6:
\[ n^2 = 49 \]

### Step 8: Solve for \( n \)
Taking the square root of both sides:
\[ n = 7 \]

### Step 9: Find the largest number
The largest of the three consecutive numbers is:
\[ n + 1 = 7 + 1 = 8 \]

Thus, the largest of the three numbers is **8**.

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