Radha was asked to find the average of N consecutive natural numbers starting from 1. By mistake, he added a number twice but he didn't notice it. As a result, he obtained a wrong average of `45(11)/(18)`.. Find the number he added twice.

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the Problem
Radha was supposed to find the average of N consecutive natural numbers starting from 1. However, he mistakenly added one number (let's call it x) twice, leading to a wrong average of \( 45 \frac{11}{18} \).

### Step 2: Convert the Wrong Average to an Improper Fraction
First, we convert the mixed number \( 45 \frac{11}{18} \) into an improper fraction:
\[ 45 \frac{11}{18} = \frac{45 \times 18 + 11}{18} = \frac{810 + 11}{18} = \frac{821}{18} \]

### Step 3: Set Up the Equation for the Average
The average of the first N natural numbers (1 to N) is given by:
\[ \text{Average} = \frac{\text{Sum of first N natural numbers}}{N} = \frac{\frac{N(N + 1)}{2}}{N} = \frac{N + 1}{2} \]

When Radha added the number x twice, the new average becomes:
\[ \text{New Average} = \frac{\frac{N(N + 1)}{2} + x}{N + 1} \]
Setting this equal to the wrong average:
\[ \frac{\frac{N(N + 1)}{2} + x}{N + 1} = \frac{821}{18} \]

### Step 4: Cross-Multiply to Eliminate the Fraction
Cross-multiplying gives:
\[ 18\left(\frac{N(N + 1)}{2} + x\right) = 821(N + 1) \]

### Step 5: Simplify the Equation
Expanding both sides:
\[ 9N(N + 1) + 18x = 821N + 821 \]
Rearranging gives:
\[ 9N^2 + 9N + 18x = 821N + 821 \]
\[ 9N^2 - 812N + 18x - 821 = 0 \]

### Step 6: Solve for x
To find x, we will check two cases for the average, as suggested in the transcript.

#### Case 1: Assume the correct average is 45
Setting:
\[ \frac{N + 1}{2} = 45 \implies N + 1 = 90 \implies N = 89 \]
Calculating the sum:
\[ \text{Sum} = \frac{89 \times 90}{2} = 4005 \]
Now, using the wrong average:
\[ \frac{4005 + x}{90} = \frac{821}{18} \]
Cross-multiplying gives:
\[ 18(4005 + x) = 821 \times 90 \]
Calculating \( 821 \times 90 = 73990 \):
\[ 72090 + 18x = 73990 \implies 18x = 73990 - 72090 = 990 \implies x = \frac{990}{18} = 55 \]

#### Case 2: Assume the correct average is 45.5
Setting:
\[ \frac{N + 1}{2} = 45.5 \implies N + 1 = 91 \implies N = 90 \]
Calculating the sum:
\[ \text{Sum} = \frac{90 \times 91}{2} = 4095 \]
Using the wrong average:
\[ \frac{4095 + x}{91} = \frac{821}{18} \]
Cross-multiplying gives:
\[ 18(4095 + x) = 821 \times 91 \]
Calculating \( 821 \times 91 = 74711 \):
\[ 73710 + 18x = 74711 \implies 18x = 74711 - 73710 = 1001 \implies x = \frac{1001}{18} \approx 55.61 \]

### Conclusion
The number that Radha added twice is \( x = 10 \).