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

Answer

To solve the problem step by step, we will use the formula for the n-th term of an arithmetic progression (A.P.), which is given by:

\[ A_n = A + (n - 1)D \]

where:
- \( A \) is the first term,
- \( D \) is the common difference,
- \( n \) is the term number.

### Step 1: Write down the equations for the given terms.

From the problem, we know:
- The 10th term \( A_{10} = -4 \)
- The 22nd term \( A_{22} = -16 \)

Using the formula for the n-th term, we can write:

1. For the 10th term:
\[ A + (10 - 1)D = -4 \implies A + 9D = -4 \quad \text{(Equation 1)} \]

2. For the 22nd term:
\[ A + (22 - 1)D = -16 \implies A + 21D = -16 \quad \text{(Equation 2)} \]

### Step 2: Subtract Equation 1 from Equation 2.

Now, we will eliminate \( A \) by subtracting Equation 1 from Equation 2:

\[ (A + 21D) - (A + 9D) = -16 - (-4) \]

This simplifies to:

\[ 21D - 9D = -16 + 4 \]

\[ 12D = -12 \]

### Step 3: Solve for \( D \).

Now, we can solve for \( D \):

\[ D = \frac{-12}{12} = -1 \]

### Step 4: Substitute \( D \) back into one of the equations to find \( A \).

Now that we have \( D \), we can substitute it back into Equation 1 to find \( A \):

\[ A + 9(-1) = -4 \]

\[ A - 9 = -4 \]

\[ A = -4 + 9 = 5 \]

### Step 5: Find the 38th term \( A_{38} \).

Now we can find the 38th term using the formula:

\[ A_{38} = A + (38 - 1)D \]

Substituting the values of \( A \) and \( D \):

\[ A_{38} = 5 + (37)(-1) \]

\[ A_{38} = 5 - 37 \]

\[ A_{38} = -32 \]

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

---