The sum of an infinite G.P. is 16 and the sum of the squares of its terms is `153(3)/(5)` . Find the fourth term of the progression :

Options

A. `7/(16)`
B. `5/(32)`
C. `3/(16)`
D. none of these

Answer

To solve the problem step by step, we will use the formulas for the sum of an infinite geometric progression (G.P.) and the sum of the squares of its terms.

### Step 1: Understand the given information
We know:
- The sum of the infinite G.P. is \( S_{\infty} = 16 \).
- The sum of the squares of its terms is \( S_{\text{squares}} = 153 \frac{3}{5} = \frac{768}{5} \).

### Step 2: Use the formula for the sum of an infinite G.P.
The formula for the sum of an infinite G.P. is given by:
\[ S_{\infty} = \frac{A}{1 - r} \]
where \( A \) is the first term and \( r \) is the common ratio.

From the information given:
\[ \frac{A}{1 - r} = 16 \quad \text{(1)} \]

### Step 3: Use the formula for the sum of the squares of the terms
The sum of the squares of the terms of a G.P. is given by:
\[ S_{\text{squares}} = \frac{A^2}{1 - r^2} \]
From the information given:
\[ \frac{A^2}{1 - r^2} = \frac{768}{5} \quad \text{(2)} \]

### Step 4: Square the first equation
Squaring equation (1):
\[ \left(\frac{A}{1 - r}\right)^2 = 16^2 = 256 \]
This gives us:
\[ \frac{A^2}{(1 - r)^2} = 256 \quad \text{(3)} \]

### Step 5: Divide equation (2) by equation (3)
Now, we will divide equation (2) by equation (3):
\[ \frac{\frac{A^2}{1 - r^2}}{\frac{A^2}{(1 - r)^2}} = \frac{\frac{768}{5}}{256} \]
This simplifies to:
\[ \frac{(1 - r)^2}{1 - r^2} = \frac{768}{5 \times 256} \]
Calculating the right side:
\[ \frac{768}{1280} = \frac{48}{80} = \frac{3}{5} \]
Thus, we have:
\[ \frac{(1 - r)^2}{(1 - r)(1 + r)} = \frac{3}{5} \]
Cancelling \( (1 - r) \) (assuming \( r \neq 1 \)):
\[ \frac{1 - r}{1 + r} = \frac{3}{5} \]

### Step 6: Cross-multiply and solve for \( r \)
Cross-multiplying gives:
\[ 5(1 - r) = 3(1 + r) \]
Expanding this:
\[ 5 - 5r = 3 + 3r \]
Rearranging terms:
\[ 5 - 3 = 5r + 3r \]
\[ 2 = 8r \]
Thus:
\[ r = \frac{2}{8} = \frac{1}{4} \]

### Step 7: Substitute \( r \) back to find \( A \)
Substituting \( r \) back into equation (1):
\[ \frac{A}{1 - \frac{1}{4}} = 16 \]
This simplifies to:
\[ \frac{A}{\frac{3}{4}} = 16 \]
Thus:
\[ A = 16 \times \frac{3}{4} = 12 \]

### Step 8: Find the fourth term of the G.P.
The fourth term of the G.P. is given by:
\[ T_4 = A r^3 \]
Substituting \( A \) and \( r \):
\[ T_4 = 12 \left(\frac{1}{4}\right)^3 = 12 \times \frac{1}{64} = \frac{12}{64} = \frac{3}{16} \]

### Final Answer
The fourth term of the progression is \( \frac{3}{16} \).

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