Find A.P., if T n' T_{m} are as given below :


T_{2} = 1 T_{12} = - 9

To find the Arithmetic Progression (A.P.) given \( T_2 = 1 \) and \( T_{12} = -9 \), we can use the general formula for the \(n\)-th term of an A.P., which is:

\[
T_n = a + (n - 1)d
\]

Where:
- \( T_n \) is the \(n\)-th term,
- \( a \) is the first term of the A.P.,
- \( d \) is the common difference,
- \( n \) is the position of the term.

### Step 1: Using the formula for \(T_2\)
For \( T_2 = 1 \):

\[
T_2 = a + (2 - 1)d = a + d
\]
So, 
\[
a + d = 1 \quad \text{(Equation 1)}
\]

### Step 2: Using the formula for \(T_{12}\)
For \( T_{12} = -9 \):

\[
T_{12} = a + (12 - 1)d = a + 11d
\]
So,
\[
a + 11d = -9 \quad \text{(Equation 2)}
\]

### Step 3: Solving the system of equations
Now, solve the two equations:
1. \( a + d = 1 \)
2. \( a + 11d = -9 \)

Subtract Equation 1 from Equation 2:
\[
(a + 11d) - (a + d) = -9 - 1
\]
\[
10d = -10
\]
\[
d = -1
\]

### Step 4: Finding the first term \(a\)
Substitute \( d = -1 \) into Equation 1:
\[
a + (-1) = 1
\]
\[
a = 2
\]

### Final Answer:
- The first term \( a = 2 \)
- The common difference \( d = -1 \)

Thus, the Arithmetic Progression (A.P.) is:

\[
T_n = 2 + (n - 1)(-1)
\]

Would you like more details or have any questions?

### Related Questions:
1. How do we calculate the sum of the first \(n\) terms of an A.P.?
2. How does the common difference affect the pattern of an A.P.?
3. What is the 20th term of this A.P.?
4. How do you derive the formula for the general term of an A.P.?
5. What is the sum of the first 12 terms of this A.P.?

### Tip:
When solving problems involving arithmetic progressions, always start by writing down the general term and identifying known values like \(T_n\) and \(T_m\) to create solvable equations.

This problem explores how to find the first term (a) and common difference (d) of an arithmetic progression (A.P.) using the given terms T_2 = 1 and T_{12} = -9. Step-by-step calculations with the general A.P. formula T_n = a + (n - 1)d are provided. Suitable for students in Grades 9-12.

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