The sum of the series `(1)/(2) + (1)/(3) + (1)/(6)+…..` to 9 terms is

Options

A. `-5//6`
B. `-1//2`
C. 1
D. `-3//2`

Answer

To find the sum of the series \( \frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \ldots \) up to 9 terms, we need to first identify the nature of the series and then apply the formula for the sum of an arithmetic progression (AP).

### Step-by-Step Solution:

1. **Identify the Series**:
The given series is \( \frac{1}{2}, \frac{1}{3}, \frac{1}{6}, \ldots \). We need to check if this series is an arithmetic progression (AP).

2. **Calculate the Differences**:
- Find the difference between the first two terms:
\[ D_1 = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6} \]
- Now, find the difference between the second and third terms:
\[ D_2 = \frac{1}{6} - \frac{1}{3} = \frac{1 - 2}{6} = -\frac{1}{6} \]
- Since \( D_1 = D_2 \), we can conclude that the series is indeed an AP.

3. **Determine the First Term and Common Difference**:
- The first term \( A = \frac{1}{2} \)
- The common difference \( D = -\frac{1}{6} \)

4. **Use the Sum Formula for AP**:
The formula for the sum of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \left(2A + (n - 1)D\right) \]
For our case, \( n = 9 \), \( A = \frac{1}{2} \), and \( D = -\frac{1}{6} \).

5. **Substitute Values into the Formula**:
\[ S_9 = \frac{9}{2} \left(2 \cdot \frac{1}{2} + (9 - 1) \cdot -\frac{1}{6}\right) \]
Simplifying this:
\[ S_9 = \frac{9}{2} \left(1 + 8 \cdot -\frac{1}{6}\right) \]
\[ S_9 = \frac{9}{2} \left(1 - \frac{8}{6}\right) \]
\[ S_9 = \frac{9}{2} \left(1 - \frac{4}{3}\right) \]
\[ S_9 = \frac{9}{2} \left(-\frac{1}{3}\right) \]

6. **Final Calculation**:
\[ S_9 = \frac{9 \cdot -1}{2 \cdot 3} = -\frac{3}{2} \]

Thus, the sum of the series up to 9 terms is \( \boxed{-\frac{3}{2}} \).