Average of 9 consecutive natural numbers is 41. What is the largest among these numbers?

Options

A. 37
B. 44
C. 46
D. 45

Answer

To solve the problem, we need to find the largest among 9 consecutive natural numbers whose average is 41.

### Step-by-Step Solution:

1. **Understanding the Average**:
The average of a set of numbers is calculated by dividing the sum of those numbers by the count of the numbers. Here, we are given that the average of 9 consecutive natural numbers is 41.

2. **Setting Up the Equation**:
Let the first of the 9 consecutive natural numbers be \( x \). Therefore, the numbers can be represented as:
\[ x, x+1, x+2, x+3, x+4, x+5, x+6, x+7, x+8 \]

3. **Calculating the Sum**:
The sum of these 9 numbers can be expressed as:
\[ \text{Sum} = x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) + (x+7) + (x+8) \]
This simplifies to:
\[ \text{Sum} = 9x + (0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 9x + 36 \]

4. **Using the Average**:
Since the average is given as 41, we can set up the equation:
\[ \frac{9x + 36}{9} = 41 \]

5. **Solving the Equation**:
To eliminate the fraction, multiply both sides by 9:
\[ 9x + 36 = 369 \]
Now, subtract 36 from both sides:
\[ 9x = 333 \]
Finally, divide by 9:
\[ x = 37 \]

6. **Finding the Largest Number**:
The largest number among the 9 consecutive natural numbers is:
\[ x + 8 = 37 + 8 = 45 \]

### Conclusion:
The largest among these numbers is **45**.