Question

If the mean of first n natural numbers is \(\frac{6n}{11}\), then n is

• A. 11
• B. 6
• C. 12
• D. 22

Hint:

The mean of consecutive terms in an A.P. is simply the average of the first and last term:
\[ \text{Mean} = \frac{1 + n}{2} \]
This simple trick bypasses writing down the sum formula!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given that the arithmetic mean of the first \(n\) natural numbers is equal to \(\frac{6n}{11}\). We need to find the value of \(n\).

Step 2: Key Formula or Approach:
- The sum of the first \(n\) natural numbers is given by:
\[ S_n = \frac{n(n + 1)}{2} \]
- The mean of first \(n\) natural numbers is:
\[ \text{Mean} = \frac{\text{Sum}}{n} = \frac{n(n + 1)}{2 \times n} = \frac{n + 1}{2} \]

Step 3: Detailed Explanation:

• Set the mean formula equal to the given expression:
\[ \frac{n + 1}{2} = \frac{6n}{11} \]

• Cross-multiply to solve the linear equation:
\[ 11(n + 1) = 2(6n) \]
\[ 11n + 11 = 12n \]

• Rearrange the equation to isolate \(n\):
\[ 12n - 11n = 11 \]
\[ n = 11 \]


Step 4: Final Answer:
The value of \(n\) is 11.