In an increasing geometric series, the sum of the first and the sixth term is `66` and the product of the second and fifth term is `128`. Then the sum of the first `6` terms of the series is:

Options

A. `127`
B. `129`
C. `126`
D. `128`

Answer

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

### Step 1: Set up the equations based on the problem statement
We are given two conditions:
1. The sum of the first and sixth terms is 66.
2. The product of the second and fifth terms is 128.

Using the formula for the \( n \)-th term of a geometric series, we can express these conditions mathematically.

- The first term is \( a \).
- The sixth term is \( a r^5 \).
- The second term is \( a r \).
- The fifth term is \( a r^4 \).

From the first condition:
\[ a + a r^5 = 66 \quad \text{(Equation 1)} \]

From the second condition:
\[ (a r)(a r^4) = 128 \implies a^2 r^5 = 128 \quad \text{(Equation 2)} \]

### Step 2: Express \( a r^5 \) from Equation 1
From Equation 1, we can express \( a r^5 \):
\[ a r^5 = 66 - a \]

### Step 3: Substitute \( a r^5 \) into Equation 2
Now, substitute \( a r^5 \) from above into Equation 2:
\[ a^2 (66 - a) = 128 \]

### Step 4: Expand and rearrange the equation
Expanding the equation gives:
\[ 66a^2 - a^3 = 128 \]
Rearranging gives us:
\[ a^3 - 66a^2 + 128 = 0 \quad \text{(Equation 3)} \]

### Step 5: Solve the cubic equation
We will use the factorization method to solve Equation 3. We can try to find rational roots using the Rational Root Theorem. Testing \( a = 64 \) and \( a = 2 \):

1. For \( a = 64 \):
\[ 64^3 - 66 \cdot 64^2 + 128 = 0 \]
This does not hold true.

2. For \( a = 2 \):
\[ 2^3 - 66 \cdot 2^2 + 128 = 0 \]
This holds true.

Thus, the possible values for \( a \) are \( 2 \) and \( 64 \).

### Step 6: Find the corresponding values of \( r \)
We will substitute \( a = 2 \) into Equation 2 to find \( r \):
\[ (2^2) r^5 = 128 \implies 4 r^5 = 128 \implies r^5 = 32 \implies r = 2 \]

Now, substituting \( a = 64 \):
\[ (64^2) r^5 = 128 \implies 4096 r^5 = 128 \implies r^5 = \frac{128}{4096} = \frac{1}{32} \implies r = \frac{1}{2} \]
However, since \( r \) must be greater than 1 for an increasing geometric series, we discard \( a = 64 \).

### Step 7: Calculate the sum of the first 6 terms
Now, we have \( a = 2 \) and \( r = 2 \). The formula for the sum of the first \( n \) terms of a geometric series is:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
For \( n = 6 \):
\[ S_6 = \frac{2(2^6 - 1)}{2 - 1} = \frac{2(64 - 1)}{1} = 2 \cdot 63 = 126 \]

### Final Answer
The sum of the first 6 terms of the series is \( \boxed{126} \).