Central Tendency & DispersionMTP Dec 2023 Series IIQuestion 2927 of 473
All Questions

If the arithmetic mean of 1st\displaystyle 1^{st} n\displaystyle n natural numbers is 6n11\displaystyle \frac{6n}{11} then the value of 'n\displaystyle n' is:

Options

A10
B11
C14
DNone of these
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Correct Answer

✅ Option b — 11

All Options:

  • A10
  • B11
  • C14
  • DNone of these

Detailed Solution & Explanation

**Step 1: Recall the AM of first n\displaystyle n natural numbers.** AM=n+12\text{AM} = \frac{n+1}{2} **Step 2: Set equal to given expression and solve.** n+12=6n11\frac{n+1}{2} = \frac{6n}{11} 11(n+1)=12n11(n+1) = 12n 11n+11=12n11n + 11 = 12n n=11n = 11 **Note:** The computed value is **n=11\displaystyle n = 11** (Option B), not 14 (Option C). Let us verify: AM of first 11 natural numbers =6/2=6\displaystyle = 6/2 = 6. Formula: 6×11/11=6\displaystyle 6 \times 11/11 = 6 ✓. Hence, **Option B** is the correct answer.

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