In a G.P. of real numbers, the sum of the first two terms is 7. The sum of the first six terms is 91. The sum of the first four terms is`"______________"`

Answer

To solve the problem step by step, we will denote the first term of the geometric progression (G.P.) as \( a \) and the common ratio as \( r \).

### Step 1: Set up the equations based on the given information
We know:
1. The sum of the first two terms is 7:
\[ a + ar = 7 \quad \text{(Equation 1)} \]
This can be factored as:
\[ a(1 + r) = 7 \]

2. The sum of the first six terms is 91:
\[ a + ar + ar^2 + ar^3 + ar^4 + ar^5 = 91 \]
This can be expressed using the formula for the sum of the first \( n \) terms of a G.P.:
\[ S_n = a \frac{1 - r^n}{1 - r} \quad \text{(for } r \neq 1\text{)} \]
Thus, for \( n = 6 \):
\[ S_6 = a \frac{1 - r^6}{1 - r} = 91 \quad \text{(Equation 2)} \]

### Step 2: Substitute \( a \) from Equation 1 into Equation 2
From Equation 1, we have:
\[ a = \frac{7}{1 + r} \]
Substituting this into Equation 2:
\[ \frac{7}{1 + r} \cdot \frac{1 - r^6}{1 - r} = 91 \]
Multiplying both sides by \( (1 + r)(1 - r) \):
\[ 7(1 - r^6) = 91(1 - r)(1 + r) \]
Simplifying the right side:
\[ 7(1 - r^6) = 91(1 - r^2) \]
This leads to:
\[ 7 - 7r^6 = 91 - 91r^2 \]
Rearranging gives:
\[ 7r^6 - 91r^2 + 84 = 0 \]

### Step 3: Let \( x = r^2 \) and solve the quadratic equation
We can rewrite the equation as:
\[ 7x^3 - 91x + 84 = 0 \]
Dividing the entire equation by 7:
\[ x^3 - 13x + 12 = 0 \]
Now, we can factor this cubic equation. Testing possible rational roots, we find:
\[ (x - 1)(x^2 + x - 12) = 0 \]
Factoring further:
\[ (x - 1)(x - 3)(x + 4) = 0 \]
This gives us the roots:
\[ x = 1, \quad x = 3, \quad x = -4 \]
Since \( x = r^2 \), we discard \( x = -4 \) as it is not valid for real numbers. Thus, we have:
\[ r^2 = 1 \quad \Rightarrow \quad r = 1 \quad \text{or} \quad r^2 = 3 \quad \Rightarrow \quad r = \sqrt{3} \text{ or } -\sqrt{3} \]

### Step 4: Find \( a \) for valid \( r \)
1. If \( r = 1 \):
\[ a(1 + 1) = 7 \quad \Rightarrow \quad 2a = 7 \quad \Rightarrow \quad a = \frac{7}{2} \]
The sum of the first four terms:
\[ S_4 = 4a = 4 \cdot \frac{7}{2} = 14 \]

2. If \( r = \sqrt{3} \):
\[ a(1 + \sqrt{3}) = 7 \quad \Rightarrow \quad a = \frac{7}{1 + \sqrt{3}} \]
The sum of the first four terms:
\[ S_4 = a(1 + r + r^2 + r^3) = a(1 + \sqrt{3} + 3 + 3\sqrt{3}) = a(4 + 4\sqrt{3}) \]
Substituting \( a \):
\[ S_4 = \frac{7}{1 + \sqrt{3}}(4 + 4\sqrt{3}) = 28 \]

### Final Answer
The sum of the first four terms is:
\[ \boxed{28} \]