Find the number of terms in the sequence 95, 99, 103, 107, ... 335.

Answer

To find the number of terms in the sequence 95, 99, 103, 107, ..., 335, we can follow these steps:

### Step 1: Identify the first term and common difference
The first term (a) of the sequence is 95. The common difference (d) can be calculated as:
\[ d = 99 - 95 = 4 \]

### Step 2: Use the formula for the nth term of an arithmetic progression (AP)
The formula for the nth term (T_n) of an arithmetic progression is given by:
\[ T_n = a + (n - 1) \cdot d \]
where:
- \( T_n \) is the nth term,
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the number of terms.

### Step 3: Set the nth term equal to the last term
In this case, the last term (T_n) is 335. Therefore, we can set up the equation:
\[ 335 = 95 + (n - 1) \cdot 4 \]

### Step 4: Solve for n
Now, we will solve for n:
1. Subtract 95 from both sides:
\[ 335 - 95 = (n - 1) \cdot 4 \]
\[ 240 = (n - 1) \cdot 4 \]

2. Divide both sides by 4:
\[ \frac{240}{4} = n - 1 \]
\[ 60 = n - 1 \]

3. Add 1 to both sides:
\[ n = 60 + 1 \]
\[ n = 61 \]

### Conclusion
The number of terms in the sequence is 61.

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