Three numbers in A.P are removed from first n consecutive natural numbers and average of remaining numbers is found to be `43/4` . Find n as well as removed numbers if one of the removed number is perfect square .

Answer

To solve the problem step by step, we will follow the reasoning and calculations outlined in the video transcript.

### Step 1: Define the Variables
Let the three numbers removed from the first \( n \) consecutive natural numbers be \( A - D, A, A + D \), where \( A \) is the middle term and \( D \) is the common difference.

### Step 2: Calculate the Sum of Remaining Numbers
The sum of the first \( n \) natural numbers is given by:
\[ S_n = \frac{n(n + 1)}{2} \]
After removing the three numbers, the sum of the remaining numbers \( S \) can be expressed as:
\[ S = S_n - S_3 = S_n - (A - D + A + A + D) = S_n - 3A \]
Thus, we have:
\[ S = \frac{n(n + 1)}{2} - 3A \]

### Step 3: Use the Average of Remaining Numbers
The average of the remaining numbers is given as \( \frac{43}{4} \). The number of remaining numbers is \( n - 3 \). Therefore, we can write:
\[ \frac{S}{n - 3} = \frac{43}{4} \]
Substituting for \( S \):
\[ \frac{\frac{n(n + 1)}{2} - 3A}{n - 3} = \frac{43}{4} \]
Cross-multiplying gives:
\[ 4\left(\frac{n(n + 1)}{2} - 3A\right) = 43(n - 3) \]
Simplifying this, we have:
\[ 2n(n + 1) - 12A = 43n - 129 \]
Rearranging leads to:
\[ 2n^2 - 41n + 12A + 129 = 0 \quad \text{(Equation 1)} \]

### Step 4: Ensure Real Roots
For \( n \) to be a valid number of terms, the quadratic equation must have real roots. This means the discriminant \( D \) must be non-negative:
\[ D = b^2 - 4ac = (-41)^2 - 4 \cdot 2 \cdot (12A + 129) \geq 0 \]
Calculating the discriminant:
\[ 1681 - 8(12A + 129) \geq 0 \]
This simplifies to:
\[ 1681 - 96A - 1032 \geq 0 \implies 649 - 96A \geq 0 \implies A \leq \frac{649}{96} \approx 6.77 \]

### Step 5: Finding Suitable Values for \( A \)
Since \( A \) must be a natural number and one of the removed numbers is a perfect square, we can check perfect squares \( A = 1, 4, 9, 16 \) and so on, but we are limited to \( A \leq 6 \).

### Step 6: Testing Values of \( A \)
1. **If \( A = 4 \)**:
\[ 2n^2 - 41n + 12(4) + 129 = 0 \implies 2n^2 - 41n + 165 = 0 \]
The discriminant is:
\[ D = 41^2 - 4 \cdot 2 \cdot 165 = 1681 - 1320 = 361 \quad (\text{perfect square}) \]
Solving for \( n \):
\[ n = \frac{41 \pm 19}{4} \implies n = 15 \text{ or } n = 3 \]

2. **If \( A = 6 \)**:
\[ 2n^2 - 41n + 12(6) + 129 = 0 \implies 2n^2 - 41n + 201 = 0 \]
The discriminant is:
\[ D = 41^2 - 4 \cdot 2 \cdot 201 = 1681 - 1608 = 73 \quad (\text{not a perfect square}) \]

### Conclusion
The valid values are:
- For \( A = 4 \), \( n = 15 \) (with removed numbers being \( 2, 4, 6 \) which are in A.P. and one is a perfect square).
- For \( A = 6 \), \( n = 19 \) (with removed numbers being \( 4, 6, 8 \) which are in A.P. and one is a perfect square).

### Final Answer
Thus, the values of \( n \) and the removed numbers are:
- \( n = 15 \) with removed numbers \( 2, 4, 6 \)
- \( n = 19 \) with removed numbers \( 4, 6, 8 \)