In an increasing geometric progression, the sum of the first and the last term is 99, the product of the second and the last but one term is 288 and the sum of all the terms is 189. Then, the number of terms in the progression is equal to

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the problem, we will use the properties of a geometric progression (GP) and the information provided in the question.

### Step 1: Define the terms of the GP
Let:
- The first term be \( a \)
- The common ratio be \( r \)
- The number of terms be \( n \)

The last term of the GP can be expressed as:
\[ T_n = a r^{n-1} \]

### Step 2: Use the first condition
The sum of the first and the last term is given as 99:
\[ a + a r^{n-1} = 99 \]
This can be simplified to:
\[ a(1 + r^{n-1}) = 99 \quad \text{(Equation 1)} \]

### Step 3: Use the second condition
The product of the second term and the last but one term is given as 288:
\[ (a r)(a r^{n-2}) = 288 \]
This simplifies to:
\[ a^2 r^{n-1} = 288 \quad \text{(Equation 2)} \]

### Step 4: Use the third condition
The sum of all the terms is given as 189. The formula for the sum of the first \( n \) terms of a GP is:
\[ S_n = \frac{a(r^n - 1)}{r - 1} = 189 \quad \text{(Equation 3)} \]

### Step 5: Solve the equations
From Equation 1, we can express \( r^{n-1} \) in terms of \( a \):
\[ r^{n-1} = \frac{99}{a} - 1 \quad \text{(Equation 4)} \]

Substituting Equation 4 into Equation 2:
\[ a^2 \left(\frac{99}{a} - 1\right) = 288 \]
Expanding this gives:
\[ 99a - a^2 = 288 \]
Rearranging leads to:
\[ a^2 - 99a + 288 = 0 \quad \text{(Quadratic Equation)} \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = \frac{99 \pm \sqrt{99^2 - 4 \cdot 1 \cdot 288}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 99^2 = 9801 \]
\[ 4 \cdot 288 = 1152 \]
\[ 99^2 - 4 \cdot 288 = 9801 - 1152 = 8649 \]
Now, calculate \( \sqrt{8649} = 93 \):
\[ a = \frac{99 \pm 93}{2} \]
This gives two possible values:
1. \( a = \frac{192}{2} = 96 \)
2. \( a = \frac{6}{2} = 3 \)

### Step 7: Find \( r \) and \( n \)
Using \( a = 3 \):
From Equation 4:
\[ r^{n-1} = \frac{99}{3} - 1 = 32 \implies r^{n-1} = 32 \implies r = 2 \quad \text{(since } 32 = 2^5\text{)} \]
Thus, \( n - 1 = 5 \) implies \( n = 6 \).

Using \( a = 96 \):
From Equation 4:
\[ r^{n-1} = \frac{99}{96} - 1 = \frac{3}{96} = \frac{1}{32} \implies r^{n-1} = \frac{1}{32} \implies r = \frac{1}{2} \]
Thus, \( n - 1 = -5 \) which is not valid since \( n \) must be positive.

### Conclusion
The only valid solution is when \( a = 3 \) and \( r = 2 \), leading to:
\[ \text{Number of terms } n = 6 \]

Thus, the answer is:
\[ \boxed{6} \]