The ninth term of an A.P. is equal to seven times the second term and twelth term exceeds five times the third term by 2. Find the term and the common difference.

Answer

To solve the problem, we need to find the first term (a) and the common difference (d) of the arithmetic progression (A.P.) based on the given conditions.

### Step 1: Understand the terms of an A.P.
The nth term of an A.P. can be expressed as:
\[ a_n = a + (n - 1)d \]
where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

### Step 2: Set up the equations based on the problem statement.
1. The ninth term is equal to seven times the second term:
\[ a_9 = 7a_2 \]
Using the formula for the nth term:
\[ a + 8d = 7(a + d) \]

2. The twelfth term exceeds five times the third term by 2:
\[ a_{12} = 5a_3 + 2 \]
Again using the formula for the nth term:
\[ a + 11d = 5(a + 2d) + 2 \]

### Step 3: Simplify the equations.
1. From the first equation:
\[ a + 8d = 7a + 7d \]
Rearranging gives:
\[ a + 8d - 7a - 7d = 0 \]
\[ -6a + d = 0 \]
Thus, we have:
\[ d = 6a \] (Equation 1)

2. From the second equation:
\[ a + 11d = 5a + 10d + 2 \]
Rearranging gives:
\[ a + 11d - 5a - 10d - 2 = 0 \]
\[ -4a + d - 2 = 0 \]
Thus, we have:
\[ d = 4a + 2 \] (Equation 2)

### Step 4: Solve the equations simultaneously.
Now we have two equations:
1. \( d = 6a \)
2. \( d = 4a + 2 \)

Setting them equal to each other:
\[ 6a = 4a + 2 \]
Subtract \( 4a \) from both sides:
\[ 2a = 2 \]
Dividing by 2 gives:
\[ a = 1 \]

### Step 5: Find the common difference (d).
Now substitute \( a = 1 \) into Equation 1:
\[ d = 6a = 6 \times 1 = 6 \]

### Conclusion
The first term \( a \) is 1, and the common difference \( d \) is 6.

### Final Answer
- First term (a) = 1
- Common difference (d) = 6