The `10^(th)` term of an A.P. is `-4` and its `22^(nd)` term is `(-16)`. Find its `38^(th)` term.

Answer

To find the \(38^{th}\) term of the arithmetic progression (A.P.) where the \(10^{th}\) term is \(-4\) and the \(22^{nd}\) term is \(-16\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given terms**:
- The \(10^{th}\) term, \(a_{10} = -4\)
- The \(22^{nd}\) term, \(a_{22} = -16\)

2. **Use the formula for the \(n^{th}\) term of an A.P.**:
The formula for the \(n^{th}\) term of an A.P. is given by:
\[ a_n = a + (n-1)d \]
where \(a\) is the first term and \(d\) is the common difference.

3. **Set up equations for the given terms**:
- For the \(10^{th}\) term:
\[ a + 9d = -4 \quad \text{(1)} \]
- For the \(22^{nd}\) term:
\[ a + 21d = -16 \quad \text{(2)} \]

4. **Subtract equation (1) from equation (2)**:
\[ (a + 21d) - (a + 9d) = -16 - (-4) \]
This simplifies to:
\[ 12d = -12 \]
Therefore, we find:
\[ d = -1 \]

5. **Substitute \(d\) back into equation (1) to find \(a\)**:
Using \(d = -1\) in equation (1):
\[ a + 9(-1) = -4 \]
Simplifying gives:
\[ a - 9 = -4 \]
Thus:
\[ a = -4 + 9 = 5 \]

6. **Now, find the \(38^{th}\) term**:
Using the formula for the \(n^{th}\) term again:
\[ a_{38} = a + (38-1)d = a + 37d \]
Substituting \(a = 5\) and \(d = -1\):
\[ a_{38} = 5 + 37(-1) = 5 - 37 = -32 \]

### Final Answer:
The \(38^{th}\) term of the A.P. is \(-32\).

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