Find a G.P. for which the sum of first two terms is `-4` and the fifth is 4 times the third term.

Answer

To find a geometric progression (G.P.) for which the sum of the first two terms is -4 and the fifth term is 4 times the third term, we can follow these steps:

### Step 1: Define the terms of the G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The first few terms of the G.P. can be expressed as:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)
- Fourth term: \( ar^3 \)
- Fifth term: \( ar^4 \)

### Step 2: Set up the equations based on the given conditions
From the problem, we have two conditions:
1. The sum of the first two terms is -4:
\[ a + ar = -4 \quad \text{(Equation 1)} \]

2. The fifth term is 4 times the third term:
\[ ar^4 = 4(ar^2) \quad \text{(Equation 2)} \]

### Step 3: Simplify Equation 2
From Equation 2, we can simplify:
\[ ar^4 = 4ar^2 \]
Assuming \( a \neq 0 \) (since if \( a = 0 \), all terms would be zero), we can divide both sides by \( a \):
\[ r^4 = 4r^2 \]
Now, we can rearrange this equation:
\[ r^4 - 4r^2 = 0 \]
Factoring out \( r^2 \):
\[ r^2(r^2 - 4) = 0 \]
This gives us:
\[ r^2 = 0 \quad \text{or} \quad r^2 - 4 = 0 \]
Thus, \( r^2 = 4 \) or \( r = 0 \) (not valid for G.P.).

So, we have:
\[ r = 2 \quad \text{or} \quad r = -2 \]

### Step 4: Solve for \( a \) using Equation 1
**Case 1: \( r = 2 \)**
Substituting \( r = 2 \) into Equation 1:
\[ a + 2a = -4 \]
\[ 3a = -4 \implies a = -\frac{4}{3} \]
Thus, the G.P. is:
\[ -\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \ldots \]

**Case 2: \( r = -2 \)**
Substituting \( r = -2 \) into Equation 1:
\[ a - 2a = -4 \]
\[ -a = -4 \implies a = 4 \]
Thus, the G.P. is:
\[ 4, -8, 16, \ldots \]

### Final G.P. Solutions
The two possible G.P.s that satisfy the conditions are:
1. \( -\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \ldots \) (for \( r = 2 \))
2. \( 4, -8, 16, \ldots \) (for \( r = -2 \))