The sum of an infinity decreasing G.P is equal to 4 and the sum of the cubes of its terms is equal to 64/7. Then `5^(th)` term of the progression is :

Options

A. `1/4`
B. `1/8`
C. `1/16`
D. `1/32`

Explanation

To solve the problem, we need to find the fifth term of an infinitely decreasing geometric progression (G.P.) given that the sum of the G.P. is 4 and the sum of the cubes of its terms is \( \frac{64}{7} \).

### Step-by-step Solution:

1. **Identify the formula for the sum of an infinite G.P.**
The sum \( S \) of an infinite G.P. with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by:
\[ S = \frac{a}{1 - r} \]
Given that the sum is 4, we can write:
\[ \frac{a}{1 - r} = 4 \quad \text{(1)} \]

2. **Identify the formula for the sum of the cubes of the terms of the G.P.**
The sum of the cubes of the terms of the G.P. can be expressed as another G.P. with first term \( a^3 \) and common ratio \( r^3 \):
\[ S_{\text{cubes}} = \frac{a^3}{1 - r^3} \]
Given that this sum is \( \frac{64}{7} \), we can write:
\[ \frac{a^3}{1 - r^3} = \frac{64}{7} \quad \text{(2)} \]

3. **Express \( a \) in terms of \( r \) using equation (1)**
From equation (1):
\[ a = 4(1 - r) \]

4. **Substitute \( a \) in equation (2)**
Substitute \( a \) from the previous step into equation (2):
\[ \frac{(4(1 - r))^3}{1 - r^3} = \frac{64}{7} \]
Simplifying the left side:
\[ \frac{64(1 - r)^3}{1 - r^3} = \frac{64}{7} \]
Cancel \( 64 \) from both sides:
\[ \frac{(1 - r)^3}{1 - r^3} = \frac{1}{7} \]

5. **Cross-multiply to eliminate the fraction**
Cross-multiplying gives:
\[ 7(1 - r)^3 = 1 - r^3 \]

6. **Expand and simplify**
Expand \( (1 - r)^3 \):
\[ (1 - r)^3 = 1 - 3r + 3r^2 - r^3 \]
Thus:
\[ 7(1 - 3r + 3r^2 - r^3) = 1 - r^3 \]
This simplifies to:
\[ 7 - 21r + 21r^2 - 7r^3 = 1 - r^3 \]
Rearranging gives:
\[ 6r^3 + 21r^2 - 21r + 6 = 0 \]
Dividing the entire equation by 3:
\[ 2r^3 + 7r^2 - 7r + 2 = 0 \]

7. **Factor the cubic equation**
We can try to factor this cubic equation. After testing possible rational roots, we find:
\[ (r - 1)(2r^2 + 9r - 2) = 0 \]
Solving \( 2r^2 + 9r - 2 = 0 \) using the quadratic formula:
\[ r = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{-9 \pm \sqrt{81 + 16}}{4} = \frac{-9 \pm \sqrt{97}}{4} \]

Since \( r \) must be less than 1 for a decreasing G.P., we take \( r = \frac{1}{2} \).

8. **Find \( a \)**
Substitute \( r = \frac{1}{2} \) back into equation (1):
\[ a = 4(1 - \frac{1}{2}) = 4 \cdot \frac{1}{2} = 2 \]

9. **Calculate the fifth term**
The \( n \)-th term of a G.P. is given by:
\[ a_n = a \cdot r^{n-1} \]
For the fifth term (\( n = 5 \)):
\[ a_5 = 2 \cdot \left(\frac{1}{2}\right)^{4} = 2 \cdot \frac{1}{16} = \frac{2}{16} = \frac{1}{8} \]

### Final Answer:
The fifth term of the progression is \( \frac{1}{8} \).