If the number of terms of an A.P. is `(2n+1)`, then what is the ratio of the sum of the odd terms to the sum of even terms ?

Options

A. `(n)/(n+1)`
B. `(n^(2))/(n+1)`
C. `(n+1)/(n)`
D. `(n+1)/(2n)`

Answer

To solve the problem of finding the ratio of the sum of the odd terms to the sum of the even terms in an arithmetic progression (A.P.) with a total of \(2n + 1\) terms, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the A.P. Terms**:
Let the first term of the A.P. be \(a\) and the common difference be \(d\). The terms of the A.P. can be expressed as:
\[ a, \quad a + d, \quad a + 2d, \quad a + 3d, \quad \ldots, \quad a + 2nd \]
There are \(2n + 1\) terms in total.

2. **Separate Odd and Even Terms**:
- **Odd Terms**: The odd-positioned terms are:
\[ a, \quad a + 2d, \quad a + 4d, \quad \ldots, \quad a + 2nd \]
The number of odd terms is \(n + 1\).
- **Even Terms**: The even-positioned terms are:
\[ a + d, \quad a + 3d, \quad a + 5d, \quad \ldots, \quad a + (2n - 1)d \]
The number of even terms is \(n\).

3. **Sum of Odd Terms**:
The sum of the odd terms can be calculated using the formula for the sum of an A.P.:
\[ S_{\text{odd}} = \frac{\text{Number of terms}}{2} \times (\text{First term} + \text{Last term}) \]
Here, the first term is \(a\) and the last term is \(a + 2nd\):
\[ S_{\text{odd}} = \frac{n + 1}{2} \times (a + (a + 2nd)) = \frac{n + 1}{2} \times (2a + 2nd) = (n + 1)(a + nd) \]

4. **Sum of Even Terms**:
Similarly, the sum of the even terms is:
\[ S_{\text{even}} = \frac{n}{2} \times (\text{First term} + \text{Last term}) \]
The first term is \(a + d\) and the last term is \(a + (2n - 1)d\):
\[ S_{\text{even}} = \frac{n}{2} \times ((a + d) + (a + (2n - 1)d)) = \frac{n}{2} \times (2a + (2n)d) = n(a + nd) \]

5. **Ratio of the Sums**:
Now, we can find the ratio of the sum of odd terms to the sum of even terms:
\[ \text{Ratio} = \frac{S_{\text{odd}}}{S_{\text{even}}} = \frac{(n + 1)(a + nd)}{n(a + nd)} \]
Since \(a + nd\) is common in both the numerator and denominator, we can cancel it out (assuming \(a + nd \neq 0\)):
\[ \text{Ratio} = \frac{n + 1}{n} \]

6. **Final Result**:
Therefore, the required ratio of the sum of the odd terms to the sum of the even terms is:
\[ \frac{n + 1}{n} \]