In a n increasing G.P. , the sum of the first and the last term is 66, the product of the second and the last but one is 128 and the sum of the terms is 126. How many terms are there in the progression?

Options

A. `r=1//3`
B. `r= 2 sqrt(2)//3`
C. Sum of infinite terms is 6
D. none of these

Explanation

To solve the problem step by step, let's denote the first term of the geometric progression (G.P.) as \( a \) and the common ratio as \( r \). The number of terms in the G.P. is denoted by \( n \).

### Step 1: Set up the equations based on the problem statement

1. The sum of the first and the last term is given by:
\[ a + ar^{n-1} = 66 \quad \text{(1)} \]

2. The product of the second term and the last but one term is given by:
\[ ar \cdot ar^{n-2} = 128 \quad \text{(2)} \]

3. The sum of all terms in the G.P. is given by:
\[ \frac{a(r^n - 1)}{r - 1} = 126 \quad \text{(3)} \]

### Step 2: Simplify the equations

From equation (1):
\[ a(1 + r^{n-1}) = 66 \quad \text{(4)} \]

From equation (2):
\[ a^2 r^{n-1} = 128 \quad \text{(5)} \]

### Step 3: Solve for \( a \) in terms of \( r \) and \( n \)

From equation (4):
\[ a = \frac{66}{1 + r^{n-1}} \quad \text{(6)} \]

Substituting equation (6) into equation (5):
\[ \left(\frac{66}{1 + r^{n-1}}\right)^2 r^{n-1} = 128 \]

### Step 4: Cross-multiply and simplify

Cross-multiplying gives:
\[ 66^2 r^{n-1} = 128(1 + r^{n-1})^2 \]

Expanding the right side:
\[ 66^2 r^{n-1} = 128(1 + 2r^{n-1} + r^{2(n-1)}) \]

### Step 5: Rearranging the equation

Rearranging gives:
\[ 66^2 r^{n-1} = 128 + 256 r^{n-1} + 128 r^{2(n-1)} \]

This can be rearranged to form a quadratic equation in \( r^{n-1} \):
\[ 128 r^{2(n-1)} + (256 - 66^2) r^{n-1} + 128 = 0 \]

### Step 6: Solve the quadratic equation

Calculating \( 66^2 \):
\[ 66^2 = 4356 \]
Thus, the equation becomes:
\[ 128 r^{2(n-1)} + (256 - 4356) r^{n-1} + 128 = 0 \]
\[ 128 r^{2(n-1)} - 4100 r^{n-1} + 128 = 0 \]

### Step 7: Use the quadratic formula

Using the quadratic formula \( r^{n-1} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r^{n-1} = \frac{4100 \pm \sqrt{(-4100)^2 - 4 \cdot 128 \cdot 128}}{2 \cdot 128} \]

### Step 8: Calculate the discriminant

Calculating the discriminant:
\[ 4100^2 - 4 \cdot 128 \cdot 128 = 16861000 - 65536 = 16845464 \]

### Step 9: Find the roots

Calculating the roots gives us the values of \( r^{n-1} \). After solving, we find that:
\[ r^{n-1} = 32 \quad \text{(since we are looking for increasing G.P.)} \]

### Step 10: Solve for \( n \)

Since \( r^{n-1} = 32 \), we can express \( r \) as:
\[ r = 2 \quad \text{(as it is the only valid solution for increasing G.P.)} \]

Now substituting \( r = 2 \) back into the equation:
\[ 2^{n-1} = 32 \implies n - 1 = 5 \implies n = 6 \]

### Final Answer
Thus, the number of terms in the progression is \( n = 6 \).

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