Is the sum of m terms of an AP always less than the sum of (m +1) terms? Give reason.

Answer

To determine whether the sum of \( m \) terms of an arithmetic progression (AP) is always less than the sum of \( (m + 1) \) terms, we can analyze the sums of the two series.

### Step-by-Step Solution:

1. **Formula for the Sum of n Terms of an AP**:
The sum \( S_n \) of the first \( n \) terms of an AP can be given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Sum of m Terms**:
For \( m \) terms, the sum \( S_m \) is:
\[ S_m = \frac{m}{2} \times (2a + (m - 1)d) \]

3. **Sum of (m + 1) Terms**:
For \( (m + 1) \) terms, the sum \( S_{m+1} \) is:
\[ S_{m+1} = \frac{m + 1}{2} \times (2a + md) \]

4. **Comparing \( S_m \) and \( S_{m+1} \)**:
We need to check if \( S_m < S_{m+1} \):
\[ S_m = \frac{m}{2} \times (2a + (m - 1)d) \]
\[ S_{m+1} = \frac{m + 1}{2} \times (2a + md) \]

5. **Setting Up the Inequality**:
To compare \( S_m \) and \( S_{m+1} \), we can set up the inequality:
\[ \frac{m}{2} \times (2a + (m - 1)d) < \frac{m + 1}{2} \times (2a + md) \]

6. **Simplifying the Inequality**:
Cancel out \( \frac{1}{2} \) from both sides:
\[ m \times (2a + (m - 1)d) < (m + 1) \times (2a + md) \]
Expanding both sides gives:
\[ 2am + m(m - 1)d < 2am + 2a + md + m \cdot d \]

7. **Further Simplifying**:
Cancel \( 2am \) from both sides:
\[ m(m - 1)d < 2a + md \]
Rearranging gives:
\[ m(m - 1)d - md < 2a \]
This simplifies to:
\[ m(m - 2)d < 2a \]

8. **Conclusion**:
The inequality \( m(m - 2)d < 2a \) shows that whether \( S_m < S_{m+1} \) depends on the values of \( m \), \( a \), and \( d \). If \( d \) is negative and \( m \) is large enough, \( S_m \) can be greater than \( S_{m+1} \). Therefore, the sum of \( m \) terms of an AP is **not always less than** the sum of \( (m + 1) \) terms.