In a geometric progression, the third term exceeds the second by 6 and the second exceeds the first by 9. Find (i) the first term, (ii) the common ratio and (iii) the sum of the first ten terms.

Answer

To solve the problem step by step, we will use the properties of a geometric progression (GP).

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

### Step 2: Set up the equations based on the problem statement
From the problem, we have two conditions:
1. The third term exceeds the second by 6:
\[ ar^2 = ar + 6 \quad \text{(Equation 1)} \]
2. The second term exceeds the first by 9:
\[ ar = a + 9 \quad \text{(Equation 2)} \]

### Step 3: Rearrange the equations
From Equation 1:
\[ ar^2 - ar - 6 = 0 \]
From Equation 2:
\[ ar - a - 9 = 0 \implies a(r - 1) = 9 \implies a = \frac{9}{r - 1} \quad \text{(Equation 3)} \]

### Step 4: Substitute Equation 3 into Equation 1
Substituting \( a \) from Equation 3 into Equation 1:
\[ \left(\frac{9}{r - 1}\right)r^2 - \left(\frac{9}{r - 1}\right)r - 6 = 0 \]
Multiply through by \( (r - 1) \) to eliminate the fraction:
\[ 9r^2 - 9r - 6(r - 1) = 0 \]
Expanding gives:
\[ 9r^2 - 9r - 6r + 6 = 0 \implies 9r^2 - 15r + 6 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 9, b = -15, c = 6 \):
\[ b^2 - 4ac = (-15)^2 - 4 \cdot 9 \cdot 6 = 225 - 216 = 9 \]
Thus,
\[ r = \frac{15 \pm 3}{18} \]
Calculating the two possible values for \( r \):
1. \( r = \frac{18}{18} = 1 \)
2. \( r = \frac{12}{18} = \frac{2}{3} \)

Since \( r = 1 \) does not satisfy the conditions of the problem (the terms would not be distinct), we take:
\[ r = \frac{2}{3} \]

### Step 6: Substitute \( r \) back to find \( a \)
Using Equation 3:
\[ a = \frac{9}{\frac{2}{3} - 1} = \frac{9}{\frac{2}{3} - \frac{3}{3}} = \frac{9}{-\frac{1}{3}} = -27 \]

### Step 7: Calculate the sum of the first 10 terms
The formula for the sum of the first \( n \) terms of a GP is:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
Substituting \( a = -27 \), \( r = \frac{2}{3} \), and \( n = 10 \):
\[ S_{10} = -27 \frac{1 - \left(\frac{2}{3}\right)^{10}}{1 - \frac{2}{3}} = -27 \frac{1 - \left(\frac{2}{3}\right)^{10}}{\frac{1}{3}} = -81 \left(1 - \left(\frac{2}{3}\right)^{10}\right) \]

### Final Answers
1. First term \( a = -27 \)
2. Common ratio \( r = \frac{2}{3} \)
3. Sum of the first ten terms \( S_{10} = -81 \left(1 - \left(\frac{2}{3}\right)^{10}\right) \)