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: Recall the mean formula for the first n natural numbers.
The mean of $1, 2, \dots, n$ is $\frac{n+1}{2}$.
Step 2: Test the given options instead of solving the equation.
Try $n=11$: mean $=\frac{11+1}{2}=6$, and $\frac{6n}{11} = \frac{66}{11}=6$. Both sides match.
Step 3: Rule out the rest.
Try $n=6$: mean $=3.5$ but $\frac{6(6)}{11} \approx 3.27$, which does not match, so the other options fail similarly.
\[ \boxed{n = 11} \]