Two consecutive numbers from 1,2,3 …., n are removed.The arithmetic mean of the remaining numbers is 105/4 .
The removed numbers

Options

A. a-b=d-c
B. e=0
C. `a,b-2//3,c-1 " are in " in A.P`
D. `(b+d)//a` is an integer

Explanation

To solve the problem of finding the two consecutive numbers removed from the series 1, 2, 3, ..., n, given that the arithmetic mean of the remaining numbers is \( \frac{105}{4} \), we can follow these steps:

### Step 1: Define the Removed Numbers
Let the two consecutive numbers removed be \( a \) and \( a + 1 \).

### Step 2: Calculate the Sum of the First n Natural Numbers
The sum of the first \( n \) natural numbers is given by the formula:
\[ S_n = \frac{n(n + 1)}{2} \]

### Step 3: Calculate the Sum of Remaining Numbers
When we remove \( a \) and \( a + 1 \), the sum of the remaining numbers becomes:
\[ S_{\text{remaining}} = S_n - (a + (a + 1)) = \frac{n(n + 1)}{2} - (2a + 1) \]

### Step 4: Count the Remaining Numbers
The total count of numbers remaining after removing two numbers is:
\[ n - 2 \]

### Step 5: Set Up the Equation for Arithmetic Mean
The arithmetic mean of the remaining numbers is given by:
\[ \text{Mean} = \frac{S_{\text{remaining}}}{n - 2} \]
Substituting the values we have:
\[ \frac{\frac{n(n + 1)}{2} - (2a + 1)}{n - 2} = \frac{105}{4} \]

### Step 6: Cross Multiply to Eliminate the Fraction
Cross multiplying gives us:
\[ 4\left(\frac{n(n + 1)}{2} - (2a + 1)\right) = 105(n - 2) \]
This simplifies to:
\[ 2n(n + 1) - 4(2a + 1) = 105n - 210 \]

### Step 7: Rearrange the Equation
Rearranging the equation, we get:
\[ 2n^2 + 2n - 105n + 210 - 8a - 4 = 0 \]
This simplifies to:
\[ 2n^2 - 103n + 206 - 8a = 0 \]

### Step 8: Solve for n in terms of a
We can express \( n \) in terms of \( a \):
\[ 2n^2 - 103n + (206 - 8a) = 0 \]
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ n = \frac{103 \pm \sqrt{(-103)^2 - 4 \cdot 2 \cdot (206 - 8a)}}{2 \cdot 2} \]

### Step 9: Analyze the Discriminant
The discriminant \( D \) must be a perfect square for \( n \) to be an integer:
\[ D = 10609 - 16(206 - 8a) \]
This must be a perfect square.

### Step 10: Find Suitable Values for a
By testing integer values for \( a \) and checking if \( D \) is a perfect square, we find that \( a = 7 \) and \( a + 1 = 8 \) works.

### Conclusion
The two consecutive numbers removed are \( 7 \) and \( 8 \).

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