If sum of all terms of an infinite geometric progression is $(\frac{1}{5})$ times the sum of its odd terms, then common ratio is:

Question

Accepted Answer

To solve this problem, we need to analyze the given conditions about the infinite geometric progression (GP) and its odd terms.

### Explanation:

1. **Sum of an Infinite GP:**
   The sum $ S $ of an infinite geometric progression with first term $ a $ and common ratio $ r $ (where $ |r| < 1 $) is given by:
   $$
   S = \frac{a}{1 - r}
   $$

2. **Sum of Odd Terms:**
   The odd terms of the GP form another geometric progression with the same first term $ a $ and common ratio $ r^2 $. The sum $ S_{\text{odd}} $ of this new GP is:
   $$
   S_{\text{odd}} = \frac{a}{1 - r^2}
   $$

3. **Given Condition:**
   According to the problem, the sum of all terms is $\frac{1}{5}$ times the sum of the odd terms:
   $$
   \frac{a}{1 - r} = \frac{1}{5} \times \frac{a}{1 - r^2}
   $$

4. **Simplifying the Equation:**
   Cancel $ a $ from both sides (assuming $ a \neq 0 $):
   $$
   \frac{1}{1 - r} = \frac{1}{5(1 - r^2)}
   $$

Cross-multiply to solve for $ r $:
   $$
   5(1 - r^2) = 1 - r
   $$

Expand and rearrange:
   $$
   5 - 5r^2 = 1 - r
   $$
   $$
   5r^2 - r - 4 = 0
   $$

5. **Solving the Quadratic Equation:**
   The quadratic equation $ 5r^2 - r - 4 = 0 $ can be solved using the quadratic formula $ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 5 $, $ b = -1 $, and $ c = -4 $.

$$
   r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 5 \times (-4)}}{2 \times 5}
   $$
   $$
   r = \frac{1 \pm \sqrt{1 + 80}}{10}
   $$
   $$
   r = \frac{1 \pm \sqrt{81}}{10}
   $$
   $$
   r = \frac{1 \pm 9}{10}
   $$

This gives two solutions:
   $$
   r = \frac{10}{10} = 1 \quad \text{(not valid as } |r| < 1\text{)}
   $$
   $$
   r = \frac{-8}{10} = \frac{-4}{5}
   $$

6. **Conclusion:**
   The only valid solution for the common ratio $ r $ that satisfies the condition $ |r| < 1 $ is $ r = \frac{-4}{5} $.

### Why Other Options are Incorrect:

- **Option 1: $\frac{1}{2}$**
  - Substituting $ r = \frac{1}{2} $ into the equation $ 5r^2 - r - 4 = 0 $ does not satisfy the equation.

- **Option 2: $\frac{-1}{3}$**
  - Substituting $ r = \frac{-1}{3} $ into the equation $ 5r^2 - r - 4 = 0 $ does not satisfy the equation.

- **Option 4: $\frac{1}{5}$**
  - Substituting $ r = \frac{1}{5} $ into the equation $ 5r^2 - r - 4 = 0 $ does not satisfy the equation.

Thus, the correct answer is indeed Option 3: $\frac{-4}{5}$.