Central Tendency & DispersionMTP Dec 23 - Series IQuestion 3022 of 473
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The Harmonic mean H of two numbers is 4\displaystyle 4 and their arithmetic means A\displaystyle A and the geometric mean G\displaystyle G satisfy eq. 2A+G2=27\displaystyle 2A + G^2 = 27, the numbers are

Options

A(1,3)\displaystyle (1,3)
B(9,3)\displaystyle (9,3)
C(6,3)\displaystyle (6,3)
D(4.5,7)\displaystyle (4.5, 7)
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Correct Answer

✅ Option c — (6,3)\displaystyle (6,3)

All Options:

  • A(1,3)\displaystyle (1,3)
  • B(9,3)\displaystyle (9,3)
  • C(6,3)\displaystyle (6,3)
  • D(4.5,7)\displaystyle (4.5, 7)

Detailed Solution & Explanation

**Step 1: Let the two numbers be a\displaystyle a and b\displaystyle b.** Using the relation AM×HM=GM2\displaystyle \text{AM} \times \text{HM} = \text{GM}^2 (for two numbers): A×H=G2⇒G2=4AA \times H = G^2 \Rightarrow G^2 = 4A **Step 2: Substitute into the given equation.** 2A+G2=272A + G^2 = 27 2A+4A=272A + 4A = 27 6A=276A = 27 A=4.5A = 4.5 **Step 3: Find G2\displaystyle G^2.** G2=4A=4×4.5=18G^2 = 4A = 4 \times 4.5 = 18 G=18=32G = \sqrt{18} = 3\sqrt{2} **Step 4: Find a+b\displaystyle a + b and ab\displaystyle ab.** a+b=2A=9a + b = 2A = 9 ab=G2=18ab = G^2 = 18 **Step 5: Solve the quadratic.** t2−9t+18=0t^2 - 9t + 18 = 0 (t−3)(t−6)=0(t-3)(t-6) = 0 t=3 or t=6t = 3 \text{ or } t = 6 So the numbers are **6 and 3**. **Verification with HM:** H=2aba+b=2×189=369=4✓H = \frac{2ab}{a+b} = \frac{2 \times 18}{9} = \frac{36}{9} = 4 \checkmark Hence, **Option C** is the correct answer.

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