Show that `a _(1), a _(2), …., a _(n),…` from an A.P. where `a _(n)` is defined as below :
`a _(n) = 9 - 5n`

Answer

To show that the sequence \( a_1, a_2, \ldots, a_n, \ldots \) defined by \( a_n = 9 - 5n \) forms 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 of the sequence:**
- The \( n \)-th term of the sequence is given by:
\[ a_n = 9 - 5n \]

2. **Calculate the first few terms:**
- For \( n = 1 \):
\[ a_1 = 9 - 5(1) = 9 - 5 = 4 \]
- For \( n = 2 \):
\[ a_2 = 9 - 5(2) = 9 - 10 = -1 \]
- For \( n = 3 \):
\[ a_3 = 9 - 5(3) = 9 - 15 = -6 \]
- For \( n = 4 \):
\[ a_4 = 9 - 5(4) = 9 - 20 = -11 \]

3. **List the calculated terms:**
- The first four terms are:
\[ a_1 = 4, \quad a_2 = -1, \quad a_3 = -6, \quad a_4 = -11 \]

4. **Calculate the common difference:**
- The common difference \( d \) is calculated as follows:
- Between \( a_2 \) and \( a_1 \):
\[ d = a_2 - a_1 = -1 - 4 = -5 \]
- Between \( a_3 \) and \( a_2 \):
\[ d = a_3 - a_2 = -6 - (-1) = -6 + 1 = -5 \]
- Between \( a_4 \) and \( a_3 \):
\[ d = a_4 - a_3 = -11 - (-6) = -11 + 6 = -5 \]

5. **Conclusion:**
- Since the common difference \( d \) is the same for all consecutive terms (i.e., \( d = -5 \)), we conclude that the sequence \( a_1, a_2, a_3, \ldots \) forms an Arithmetic Progression.