The 2nd and 5th terms of a GP are `(-1)/2` and `1/16` respectively. Find the sum of the GP up to 8 terms.

Answer

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

### Given:
- The 2nd term \( a_2 = -\frac{1}{2} \)
- The 5th term \( a_5 = \frac{1}{16} \)

### Step 1: Express the terms in terms of \( a_1 \) and \( r \)
In a GP, the \( n \)-th term can be expressed as:
\[ a_n = a_1 \cdot r^{n-1} \]
Thus, we can write:
- For the 2nd term:
\[ a_2 = a_1 \cdot r^{2-1} = a_1 \cdot r \]
So,
\[ a_1 \cdot r = -\frac{1}{2} \quad \text{(1)} \]

- For the 5th term:
\[ a_5 = a_1 \cdot r^{5-1} = a_1 \cdot r^4 \]
So,
\[ a_1 \cdot r^4 = \frac{1}{16} \quad \text{(2)} \]

### Step 2: Divide the equations to eliminate \( a_1 \)
Now, we can divide equation (2) by equation (1):
\[ \frac{a_1 \cdot r^4}{a_1 \cdot r} = \frac{\frac{1}{16}}{-\frac{1}{2}} \]
This simplifies to:
\[ r^3 = \frac{1}{16} \cdot -2 = -\frac{1}{8} \]

### Step 3: Solve for \( r \)
Taking the cube root:
\[ r = \sqrt[3]{-\frac{1}{8}} = -\frac{1}{2} \]

### Step 4: Substitute \( r \) back to find \( a_1 \)
Now substitute \( r \) back into equation (1):
\[ a_1 \cdot \left(-\frac{1}{2}\right) = -\frac{1}{2} \]
This gives:
\[ a_1 = 1 \]

### Step 5: Find the sum of the first 8 terms of the GP
The sum of the first \( n \) terms of a GP is given by:
\[ S_n = \frac{a_1(1 - r^n)}{1 - r} \quad \text{for } r \neq 1 \]
Substituting \( a_1 = 1 \), \( r = -\frac{1}{2} \), and \( n = 8 \):
\[ S_8 = \frac{1 \left( 1 - \left(-\frac{1}{2}\right)^8 \right)}{1 - \left(-\frac{1}{2}\right)} \]

Calculating \( \left(-\frac{1}{2}\right)^8 \):
\[ \left(-\frac{1}{2}\right)^8 = \frac{1}{256} \]
Thus,
\[ S_8 = \frac{1 \left( 1 - \frac{1}{256} \right)}{1 + \frac{1}{2}} = \frac{1 \cdot \frac{255}{256}}{\frac{3}{2}} = \frac{255}{256} \cdot \frac{2}{3} = \frac{510}{768} = \frac{85}{128} \]

### Final Answer:
The sum of the GP up to 8 terms is:
\[ \boxed{\frac{85}{128}} \]