Show that `a _(1), a _(2), …., a _(n),…` from an A.P. where `a _(n)` is defined as below :
`a _(n) = 3 + 4n`

Answer

To show that the sequence \( a_1, a_2, \ldots, a_n, \ldots \) defined by \( a_n = 3 + 4n \) is an arithmetic progression (A.P.), we need to demonstrate that the difference between consecutive terms is constant.

### Step-by-Step Solution:

1. **Define the Terms**:
- The general term of the sequence is given by:
\[ a_n = 3 + 4n \]
- We will calculate the first three terms:
- For \( n = 1 \):
\[ a_1 = 3 + 4(1) = 3 + 4 = 7 \]
- For \( n = 2 \):
\[ a_2 = 3 + 4(2) = 3 + 8 = 11 \]
- For \( n = 3 \):
\[ a_3 = 3 + 4(3) = 3 + 12 = 15 \]

2. **Calculate the Differences**:
- Now, we will find the differences between consecutive terms:
- Calculate \( a_2 - a_1 \):
\[ a_2 - a_1 = 11 - 7 = 4 \]
- Calculate \( a_3 - a_2 \):
\[ a_3 - a_2 = 15 - 11 = 4 \]

3. **Check the Constant Difference**:
- Since both differences are equal:
\[ a_2 - a_1 = 4 \quad \text{and} \quad a_3 - a_2 = 4 \]
- We can conclude that the difference between consecutive terms is constant.

4. **Conclusion**:
- Since the difference between consecutive terms \( a_2 - a_1 \) and \( a_3 - a_2 \) is the same, we can conclude that the sequence \( a_1, a_2, a_3, \ldots \) is indeed an arithmetic progression.