Given two series:

Series I: 60, 120, 24, 48, 9.6, 19.2

Series II: 100, W, X, Y, Z, 32

Both series follow the same pattern. The roots of a quadratic equation are $ \frac{y}{w} $ and $ \frac{w - x}{16} $, where $w, x, y$ are terms from Series II. Find the quadratic equation in variable $r$ with these roots.

Question

Given two series:

Series I: 60, 120, 24, 48, 9.6, 19.2

Series II: 100, W, X, Y, Z, 32

Both series follow the same pattern. The roots of a quadratic equation are $ \frac{y}{w} $ and $ \frac{w - x}{16} $, where $w, x, y$ are terms from Series II. Find the quadratic equation in variable $r$ with these roots.

Accepted Answer

**Solution:**

1. **Identify the pattern in Series I:**
   - Series I: 60, 120, 24, 48, 9.6, 19.2
   - Observe the pattern between consecutive terms:
     - 60 to 120: multiplied by 2
     - 120 to 24: divided by 5
     - 24 to 48: multiplied by 2
     - 48 to 9.6: divided by 5
     - 9.6 to 19.2: multiplied by 2

So the pattern alternates: $ \times 2, \div 5, \times 2, \div 5, \times 2 $

2. **Apply the same pattern to Series II:**
   - Series II: 100, W, X, Y, Z, 32
   - Using the same pattern:
     - 100 to W: $ \times 2 $ → W = 100 $ \times 2 = 200 $
     - W to X: $ \div 5 $ → X = 200 $ \div 5 = 40 $
     - X to Y: $ \times 2 $ → Y = 40 $ \times 2 = 80 $
     - Y to Z: $ \div 5 $ → Z = 80 $ \div 5 = 16 $
     - Z to 32: $ \times 2 $ → Check: 16 $ \times 2 = 32 $ (correct)

3. **Values found:**
   - $ w = 200 $
   - $ x = 40 $
   - $ y = 80 $
   - $ z = 16 $

4. **Roots of the quadratic equation:**
   - First root: $ \frac{y}{w} = \frac{80}{200} = \frac{2}{5} = 0.4 $
   - Second root: $ \frac{w - x}{16} = \frac{200 - 40}{16} = \frac{160}{16} = 10 $

5. **Form the quadratic equation with roots $ \alpha = 0.4 $ and $ \beta = 10 $:**
   - Sum of roots $ S = \alpha + \beta = 0.4 + 10 = 10.4 = \frac{52}{5} $
   - Product of roots $ P = \alpha \times \beta = 0.4 \times 10 = 4 $

6. **Quadratic equation with roots $ \alpha, \beta $:**
   $$ r^2 - Sr + P = 0 $$
   $$ r^2 - \frac{52}{5}r + 4 = 0 $$

7. **Check options:**
   - Option (d) is: $r^2 - \frac{52r}{5} + 4 = 0$

**Answer:** Option (d) $r^2 - \frac{52r}{5} + 4 = 0$

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