Two consecutive number from n natural numbers `1,2,3,……,` n are removed. Arithmetic mean of the remaining numbers is `(105)/(4).`
The G.M. of the removed numbers is :

Options

A. `sqrt30`
B. `sqrt42`
C. `sqrt56`
D. `sqrt72`

Answer

To solve the problem step by step, we will follow the logical flow of the reasoning presented in the video transcript.

### Step 1: Define the Problem
We have `n` natural numbers: \(1, 2, 3, \ldots, n\). Two consecutive numbers, which we can denote as \(p\) and \(p + 1\), are removed from this set. The arithmetic mean of the remaining numbers is given as \(\frac{105}{4}\).

### 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 the Remaining Numbers
After removing the two consecutive numbers \(p\) and \(p + 1\), the sum of the remaining numbers becomes:
\[ S_{\text{remaining}} = S_n - (p + (p + 1)) = \frac{n(n + 1)}{2} - (2p + 1) \]

### Step 4: Find the Number of Remaining Numbers
The number of remaining numbers after removing 2 from \(n\) is:
\[ n - 2 \]

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

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

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

### Step 8: Analyze the Equation
This is a quadratic equation in terms of \(n\). For \(n\) and \(p\) to be integers, we need to analyze the nature of the roots of this equation.

### Step 9: Substitute \(n = 2r\)
Since \(n\) must be even for \(p\) to be an integer, let \(n = 2r\):
\[ 2(2r)^2 - 103(2r) - 8p + 206 = 0 \]
This simplifies to:
\[ 8r^2 - 206r - 8p + 206 = 0 \]

### Step 10: Solve for \(p\)
Rearranging gives:
\[ 8p = 8r^2 - 206r + 206 \]
Thus,
\[ p = r^2 - \frac{206}{8}r + \frac{206}{8} \]

### Step 11: Find Integer Solutions
To ensure \(p\) is an integer, \(1 - r\) must be divisible by 4. Let \(1 - r = 4t\), leading to:
\[ r = 1 + 4t \]
Substituting back gives:
\[ n = 2(1 + 4t) = 2 + 8t \]

### Step 12: Substitute Back to Find \(p\)
Substituting \(r\) back into the equation for \(p\):
\[ p = (1 + 4t)^2 - \frac{206}{8}(1 + 4t) + \frac{206}{8} \]

### Step 13: Calculate the GM of Removed Numbers
The two removed numbers are \(p\) and \(p + 1\). The geometric mean (GM) is given by:
\[ \text{GM} = \sqrt{p(p + 1)} \]
Substituting \(p = 7\) (from solving the quadratic), we find:
\[ \text{GM} = \sqrt{7 \cdot 8} = \sqrt{56} \]

### Final Answer
The GM of the removed numbers is:
\[ \sqrt{56} \]