The sum of an infinite G.P. is 4 and the sum of their cubes is 192. Find the first term :

Options

A. a)4
B. b)8
C. c)6
D. d)2

Answer

To solve the problem, we need to find the first term \( a \) of an infinite geometric progression (G.P.) given that the sum of the G.P. is 4 and the sum of their cubes is 192.

### Step 1: Use the formula for the sum of an infinite G.P.
The formula for the sum \( S \) of an infinite G.P. is given by:

\[ S = \frac{a}{1 - r} \]

where \( a \) is the first term and \( r \) is the common ratio. According to the problem, we have:

\[ \frac{a}{1 - r} = 4 \quad \text{(1)} \]

### Step 2: Use 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 also be expressed as:

\[ S_{\text{cubes}} = \frac{a^3}{1 - r^3} \]

According to the problem, this sum is 192:

\[ \frac{a^3}{1 - r^3} = 192 \quad \text{(2)} \]

### Step 3: Solve equation (1) for \( a \)
From equation (1), we can express \( a \) in terms of \( r \):

\[ a = 4(1 - r) \quad \text{(3)} \]

### Step 4: Substitute \( a \) into equation (2)
Now, substitute equation (3) into equation (2):

\[ \frac{(4(1 - r))^3}{1 - r^3} = 192 \]

Calculating \( (4(1 - r))^3 \):

\[ \frac{64(1 - r)^3}{1 - r^3} = 192 \]

### Step 5: Simplify the equation
Now, simplify the equation:

\[ 64(1 - r)^3 = 192(1 - r^3) \]

Dividing both sides by 64:

\[ (1 - r)^3 = 3(1 - r^3) \]

### Step 6: Expand and simplify
Expanding \( (1 - r)^3 \):

\[ 1 - 3r + 3r^2 - r^3 = 3(1 - r^3) \]

Expanding the right side:

\[ 1 - 3r + 3r^2 - r^3 = 3 - 3r^3 \]

Rearranging gives:

\[ 2r^3 + 3r^2 - 3r - 2 = 0 \quad \text{(4)} \]

### Step 7: Solve the cubic equation
Now we can solve the cubic equation (4). We can try possible rational roots, such as \( r = 1 \):

\[ 2(1)^3 + 3(1)^2 - 3(1) - 2 = 2 + 3 - 3 - 2 = 0 \]

So, \( r = 1 \) is a root. We can factor the cubic equation as:

\[ (r - 1)(2r^2 + 5r + 2) = 0 \]

Now we solve \( 2r^2 + 5r + 2 = 0 \) using the quadratic formula:

\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 - 16}}{4} = \frac{-5 \pm 3}{4} \]

This gives us:

\[ r = -\frac{1}{2} \quad \text{or} \quad r = -2 \]

Since the common ratio \( r \) must satisfy \( |r| < 1 \), we take \( r = -\frac{1}{2} \).

### Step 8: Find the first term \( a \)
Now substitute \( r \) back into equation (3) to find \( a \):

\[ a = 4(1 - (-\frac{1}{2})) = 4(1 + \frac{1}{2}) = 4 \times \frac{3}{2} = 6 \]

### Conclusion
Thus, the first term \( a \) of the infinite G.P. is:

\[ \boxed{6} \]