A boy was adding natural numbers from 1 to n. By mistake, he missed a number. If the average of the remaining numbers was 33 9/13 find the number that he missed

Let's solve this problem step by step.

### Step 1: Understand the Sum and Average of Natural Numbers from 1 to \( n \)

The sum of the first \( n \) natural numbers is given by:
\[
S_n = \frac{n(n+1)}{2}
\]
The average of these \( n \) numbers would be:
\[
\text{Average} = \frac{S_n}{n} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}
\]

### Step 2: Set up the Equation

Let’s denote the missed number by \( m \). The boy calculated the sum without including \( m \), so the sum he actually computed was:
\[
S_n - m
\]
Since he averaged the remaining \( n - 1 \) numbers and got \( 33 \frac{9}{13} \) (or \( \frac{438}{13} \)), we set up the equation:
\[
\frac{S_n - m}{n - 1} = \frac{438}{13}
\]

### Step 3: Substitute \( S_n = \frac{n(n+1)}{2} \)

Substitute the expression for \( S_n \):
\[
\frac{\frac{n(n+1)}{2} - m}{n - 1} = \frac{438}{13}
\]

### Step 4: Clear the Denominator by Cross Multiplying

Multiply both sides by \( 13(n - 1) \) to eliminate the fraction:
\[
13 \left(\frac{n(n+1)}{2} - m\right) = 438(n - 1)
\]

### Step 5: Solve for \( n \) and \( m \) Based on Additional Details

Given that the average without \( m \) is close to 33, we infer that \( n \) would likely be close to 66

A boy was adding natural numbers from 1 to n. By mistake, he missed a number. If the average of the remaining numbers was 33 9/13, find the number that he missed.

Discover how to find a missed natural number when adding numbers from 1 to n, given the average of remaining numbers as 33 9/13. This problem involves arithmetic sequences, sum and average formulas, and is suitable for students in Grades 9-11.

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