Find three consecutive natural numbers such that the sum of the first and second is 15 more than the third.

Options

A. 16
B. 17
C. 18
D. 20

Answer

To find three consecutive natural numbers such that the sum of the first and second is 15 more than the third, we can follow these steps:

### Step 1: Define the Variables
Let the first natural number be represented as \( x \). Therefore, the second natural number will be \( x + 1 \) and the third natural number will be \( x + 2 \).

### Step 2: Set Up the Equation
According to the problem, the sum of the first and second natural numbers is 15 more than the third natural number. We can write this as an equation:

\[ x + (x + 1) = (x + 2) + 15 \]

### Step 3: Simplify the Equation
Now, simplify both sides of the equation:

Left side:
\[ x + (x + 1) = 2x + 1 \]

Right side:
\[ (x + 2) + 15 = x + 17 \]

So, we have:
\[ 2x + 1 = x + 17 \]

### Step 4: Solve for \( x \)
To solve for \( x \), we will isolate \( x \) on one side of the equation. Subtract \( x \) from both sides:

\[ 2x - x + 1 = 17 \]
\[ x + 1 = 17 \]

Now, subtract 1 from both sides:

\[ x = 17 - 1 \]
\[ x = 16 \]

### Step 5: Find the Consecutive Numbers
Now that we have \( x \), we can find the three consecutive natural numbers:
- First natural number: \( x = 16 \)
- Second natural number: \( x + 1 = 16 + 1 = 17 \)
- Third natural number: \( x + 2 = 16 + 2 = 18 \)

### Conclusion
The three consecutive natural numbers are **16, 17, and 18**.

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