The `6^(th)` term of an A.P is `-10` and its `10^(th)` term is - 26 .Determine the `13^(th)` term of the A.P .

Answer

To find the 13th term of the arithmetic progression (A.P.), we start with the information given about the 6th and 10th terms.

### Step-by-step Solution:

1. **Identify the formula for the nth term of an A.P.:**
The nth term of an A.P. can be expressed as:
\[ T_n = a + (n - 1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Write the equations for the 6th and 10th terms:**
- For the 6th term (\( T_6 \)):
\[ T_6 = a + (6 - 1)d = a + 5d \]
Given \( T_6 = -10 \), we can write:
\[ a + 5d = -10 \quad \text{(1)} \]

- For the 10th term (\( T_{10} \)):
\[ T_{10} = a + (10 - 1)d = a + 9d \]
Given \( T_{10} = -26 \), we can write:
\[ a + 9d = -26 \quad \text{(2)} \]

3. **Subtract equation (1) from equation (2):**
\[ (a + 9d) - (a + 5d) = -26 - (-10) \]
Simplifying this gives:
\[ 9d - 5d = -26 + 10 \]
\[ 4d = -16 \]
Dividing both sides by 4:
\[ d = -4 \]

4. **Substitute \( d \) back into equation (1) to find \( a \):**
Substitute \( d = -4 \) into equation (1):
\[ a + 5(-4) = -10 \]
Simplifying this gives:
\[ a - 20 = -10 \]
Adding 20 to both sides:
\[ a = 10 \]

5. **Now, find the 13th term (\( T_{13} \)):**
Using the formula for the nth term:
\[ T_{13} = a + (13 - 1)d = a + 12d \]
Substitute \( a = 10 \) and \( d = -4 \):
\[ T_{13} = 10 + 12(-4) \]
Simplifying this gives:
\[ T_{13} = 10 - 48 = -38 \]

### Final Answer:
The 13th term of the A.P. is \( T_{13} = -38 \).

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