The fourth term of a G.P. is greater than the first term, which is positive, by 372. The third term is greater than the second by 60. Calculate the common ratio and the first term of the progression.

Answer

To solve the problem step by step, we will define the terms of the geometric progression (G.P.) and set up equations based on the given conditions.

### 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 terms of the G.P. can be expressed as:
- First term: \( A \)
- Second term: \( AR \)
- Third term: \( AR^2 \)
- Fourth term: \( AR^3 \)

### Step 2: Set up the equations based on the problem statement
From the problem, we have two conditions:
1. The fourth term is greater than the first term by 372:
\[ AR^3 - A = 372 \]
This can be rearranged to:
\[ A(R^3 - 1) = 372 \quad \text{(Equation 1)} \]

2. The third term is greater than the second term by 60:
\[ AR^2 - AR = 60 \]
Factoring out \( AR \), we get:
\[ AR(R - 1) = 60 \quad \text{(Equation 2)} \]

### Step 3: Solve for \( AR \) from Equation 2
From Equation 2, we can express \( AR \):
\[ AR = \frac{60}{R - 1} \]

### Step 4: Substitute \( AR \) into Equation 1
Now, substitute \( AR \) into Equation 1. First, we express \( A \):
\[ A = \frac{60}{R(R - 1)} \]
Substituting \( A \) into Equation 1 gives:
\[ \frac{60}{R(R - 1)}(R^3 - 1) = 372 \]

### Step 5: Simplify the equation
Multiply both sides by \( R(R - 1) \):
\[ 60(R^3 - 1) = 372R(R - 1) \]
Expanding both sides:
\[ 60R^3 - 60 = 372R^2 - 372R \]
Rearranging gives:
\[ 60R^3 - 372R^2 + 372R - 60 = 0 \]

### Step 6: Divide through by 12 to simplify
Dividing the entire equation by 12:
\[ 5R^3 - 31R^2 + 31R - 5 = 0 \]

### Step 7: Factor the cubic equation
We can use the Rational Root Theorem or synthetic division to find the roots. Testing \( R = 5 \):
\[ 5(5^3) - 31(5^2) + 31(5) - 5 = 0 \]
This shows \( R = 5 \) is a root. We can factor out \( (R - 5) \):
\[ 5R^3 - 31R^2 + 31R - 5 = (R - 5)(5R^2 + 4R + 1) \]

### Step 8: Solve the quadratic equation
Now we solve \( 5R^2 + 4R + 1 = 0 \) using the quadratic formula:
\[ R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 - 20}}{10} \]
Since the discriminant is negative, the only real solution is \( R = 5 \).

### Step 9: Find \( A \)
Substituting \( R = 5 \) back into Equation 2 to find \( A \):
\[ AR(5 - 1) = 60 \implies 4AR = 60 \implies AR = 15 \]
Now substituting \( R = 5 \):
\[ A \cdot 5 = 15 \implies A = 3 \]

### Final Answer
The first term \( A \) is \( 3 \) and the common ratio \( R \) is \( 5 \).