If the sum of an infinitely decreasing G.P. is 3, and the sum of the squares of its terms is `9//2`, the sum of the cubes of the terms is

Options

A. `(105)/(13)`
B. `(108)/(13)`
C. `(729)/(8)`
D. `(128)/(13)`

Answer

To solve the problem, we need to find the sum of the cubes of the terms of an infinitely decreasing geometric progression (G.P.) given that the sum of the G.P. is 3 and the sum of the squares of its terms is \( \frac{9}{2} \).

### Step-by-Step Solution:

1. **Define the G.P. Terms**:
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The terms of the G.P. can be expressed as:
\[ a, ar, ar^2, ar^3, \ldots \]

2. **Sum of the G.P.**:
The formula for the sum of an infinitely decreasing G.P. is given by:
\[ S = \frac{a}{1 - r} \]
According to the problem, this sum is equal to 3:
\[ \frac{a}{1 - r} = 3 \quad \text{(1)} \]

3. **Sum of the Squares of the Terms**:
The sum of the squares of the terms of the G.P. is:
\[ S_{\text{squares}} = a^2 + (ar)^2 + (ar^2)^2 + (ar^3)^2 + \ldots = a^2(1 + r^2 + r^4 + \ldots) \]
The series \( 1 + r^2 + r^4 + \ldots \) is also a G.P. with first term 1 and common ratio \( r^2 \):
\[ S_{\text{squares}} = \frac{a^2}{1 - r^2} \]
According to the problem, this sum is equal to \( \frac{9}{2} \):
\[ \frac{a^2}{1 - r^2} = \frac{9}{2} \quad \text{(2)} \]

4. **Solve the Equations**:
From equation (1):
\[ a = 3(1 - r) \]
Substitute \( a \) into equation (2):
\[ \frac{(3(1 - r))^2}{1 - r^2} = \frac{9}{2} \]
Simplifying this gives:
\[ \frac{9(1 - r)^2}{1 - r^2} = \frac{9}{2} \]
Canceling 9 from both sides:
\[ \frac{(1 - r)^2}{1 - r^2} = \frac{1}{2} \]
Cross-multiplying yields:
\[ 2(1 - r)^2 = 1 - r^2 \]
Expanding and rearranging:
\[ 2(1 - 2r + r^2) = 1 - r^2 \]
\[ 2 - 4r + 2r^2 = 1 - r^2 \]
\[ 3r^2 - 4r + 1 = 0 \]

5. **Solve the Quadratic Equation**:
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \]
\[ r = \frac{4 \pm \sqrt{16 - 12}}{6} \]
\[ r = \frac{4 \pm 2}{6} \]
This gives us:
\[ r = 1 \quad \text{or} \quad r = \frac{1}{3} \]
Since we need an infinitely decreasing G.P., we take \( r = \frac{1}{3} \).

6. **Find \( a \)**:
Substitute \( r \) back into equation (1):
\[ a = 3(1 - \frac{1}{3}) = 3 \cdot \frac{2}{3} = 2 \]

7. **Sum of the Cubes of the Terms**:
The sum of the cubes of the terms is:
\[ S_{\text{cubes}} = a^3 + (ar)^3 + (ar^2)^3 + \ldots = a^3(1 + r^3 + r^6 + \ldots) \]
The series \( 1 + r^3 + r^6 + \ldots \) is a G.P. with first term 1 and common ratio \( r^3 \):
\[ S_{\text{cubes}} = \frac{a^3}{1 - r^3} \]
Substituting \( a = 2 \) and \( r = \frac{1}{3} \):
\[ S_{\text{cubes}} = \frac{2^3}{1 - (\frac{1}{3})^3} = \frac{8}{1 - \frac{1}{27}} = \frac{8}{\frac{26}{27}} = 8 \cdot \frac{27}{26} = \frac{216}{26} = \frac{108}{13} \]

### Final Answer:
The sum of the cubes of the terms is \( \frac{108}{13} \).