Sum of an infinite G.P. is 2 and sum of their cubes is 24, then 5th term of the G.P. is

Options

A. `3//16`
B. `3//8`
C. `-3//8`
D. `-3//16`

Answer

To solve the problem, we will follow these steps:

### Step 1: Set up the equations based on the given information
We know that the sum of an infinite geometric progression (G.P.) is given by the formula:
\[ S_{\infty} = \frac{a}{1 - r} \]
where \(a\) is the first term and \(r\) is the common ratio. We are given that:
\[ S_{\infty} = 2 \implies \frac{a}{1 - r} = 2 \quad \text{(Equation 1)} \]

We also know that the sum of the cubes of the terms of the G.P. can be expressed as another G.P.:
\[ S_{\infty} = \frac{a^3}{1 - r^3} \]
We are given that:
\[ S_{\infty} = 24 \implies \frac{a^3}{1 - r^3} = 24 \quad \text{(Equation 2)} \]

### Step 2: Solve Equation 1 for \(a\)
From Equation 1, we can express \(a\) in terms of \(r\):
\[ a = 2(1 - r) \]

### Step 3: Substitute \(a\) into Equation 2
Now we substitute \(a\) from Step 2 into Equation 2:
\[ \frac{(2(1 - r))^3}{1 - r^3} = 24 \]
This simplifies to:
\[ \frac{8(1 - r)^3}{1 - r^3} = 24 \]

### Step 4: Simplify the equation
Cross-multiplying gives:
\[ 8(1 - r)^3 = 24(1 - r^3) \]
Dividing both sides by 8:
\[ (1 - r)^3 = 3(1 - r^3) \]

### Step 5: Expand and rearrange
Expanding both sides:
\[ (1 - 3r + 3r^2 - r^3) = 3(1 - r^3) \]
This leads to:
\[ 1 - 3r + 3r^2 - r^3 = 3 - 3r^3 \]
Rearranging gives:
\[ 2r^3 + 3r^2 - 3r - 2 = 0 \]

### Step 6: Solve the cubic equation
We can use the Rational Root Theorem or synthetic division to find the roots of the cubic equation. Testing \(r = 1\):
\[ 2(1)^3 + 3(1)^2 - 3(1) - 2 = 2 + 3 - 3 - 2 = 0 \]
So, \(r = 1\) is a root. We can factor the cubic polynomial as:
\[ 2r^3 + 3r^2 - 3r - 2 = (r - 1)(2r^2 + 5r + 2) \]

### Step 7: Factor the quadratic
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:
\[ r = -\frac{1}{2} \quad \text{or} \quad r = -2 \]

### Step 8: Find \(a\) for each \(r\)
Using \(r = -\frac{1}{2}\):
\[ a = 2(1 - (-\frac{1}{2})) = 2 \cdot \frac{3}{2} = 3 \]
Using \(r = -2\):
\[ a = 2(1 - (-2)) = 2 \cdot 3 = 6 \]

### Step 9: Find the 5th term
The 5th term of the G.P. is given by:
\[ T_5 = a \cdot r^{4} \]
For \(r = -\frac{1}{2}\):
\[ T_5 = 3 \cdot \left(-\frac{1}{2}\right)^{4} = 3 \cdot \frac{1}{16} = \frac{3}{16} \]
For \(r = -2\):
\[ T_5 = 6 \cdot (-2)^{4} = 6 \cdot 16 = 96 \]

### Conclusion
The fifth term of the G.P. can either be \(\frac{3}{16}\) or \(96\). However, since the problem specifies finding the 5th term of the G.P., we conclude that the answer is:
\[ \text{5th term of the G.P. is } \frac{3}{16} \]