Two consecutive numbers from 1, 2, 3, ..., n are removed, then arithmetic mean of the remaining numbers is `105/4` then `n/10` must be equal to

Options

A. `7^(th) " term is " 16`
B. `7^(th) " term " is 18`
C. Sum of first 10 terms is `(505)/(4)`
D. Sum of first 10 terms is `(405)/(4)`

Explanation

To solve the problem step by step, we need to find the value of \( \frac{n}{10} \) given that the arithmetic mean of the remaining numbers after removing two consecutive numbers from the set \( 1, 2, 3, \ldots, n \) is \( \frac{105}{4} \).

### Step 1: Understand the Problem
We have a sequence of numbers from 1 to \( n \). We need to remove two consecutive numbers, say \( p \) and \( p + 1 \), and then calculate the arithmetic mean of the remaining numbers.

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

### Step 3: Calculate the Sum After Removing \( p \) and \( p + 1 \)
When we remove \( p \) and \( p + 1 \), the sum of the remaining numbers becomes:
\[ S' = S - (p + (p + 1)) = \frac{n(n + 1)}{2} - (2p + 1) \]

### Step 4: Calculate the Number of Remaining Terms
After removing two numbers, the count of the remaining numbers is:
\[ n - 2 \]

### Step 5: Set Up the Equation for the Arithmetic Mean
The arithmetic mean of the remaining numbers is given by:
\[ \text{Mean} = \frac{S'}{n - 2} \]
According to the problem, this mean equals \( \frac{105}{4} \). Thus, we can set up the equation:
\[ \frac{\frac{n(n + 1)}{2} - (2p + 1)}{n - 2} = \frac{105}{4} \]

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

### Step 7: Rearranging the Equation
Rearranging gives us:
\[ 2n^2 + 2n - 105n + 210 - 8p - 4 = 0 \]
This simplifies to:
\[ 2n^2 - 103n + 206 - 8p = 0 \]

### Step 8: Solve for \( n \) and \( p \)
Since \( n \) and \( p \) are integers, we can express \( n \) in terms of \( p \):
\[ 2n^2 - 103n + (206 - 8p) = 0 \]
Using the quadratic formula:
\[ n = \frac{103 \pm \sqrt{(103)^2 - 4 \cdot 2 \cdot (206 - 8p)}}{2 \cdot 2} \]

### Step 9: Finding Integer Solutions
To ensure \( n \) is an integer, the discriminant must be a perfect square. We can analyze this further by substituting values for \( p \).

### Step 10: Testing Values
Assuming \( n \) is even, let \( n = 2k \). We can substitute and simplify to find valid integer values for \( k \) and \( p \).

### Step 11: Conclusion
After testing values, we find that \( n = 50 \) and \( p = 7 \). Thus:
\[ \frac{n}{10} = \frac{50}{10} = 5 \]

### Final Answer
The value of \( \frac{n}{10} \) is \( 5 \).