Find how many terms of the sequence 114, 121, 128, ... below 245.

Answer

To find how many terms of the sequence 114, 121, 128, ... are below 245, we can follow these steps:

### Step 1: Identify the first term and the common difference
The first term \( a \) of the sequence is 114. The common difference \( d \) can be calculated as follows:
\[ d = 121 - 114 = 7 \]

### Step 2: Write the formula for the nth term of an arithmetic sequence
The formula for the nth term \( a_n \) of an arithmetic sequence is given by:
\[ a_n = a + (n - 1) \cdot d \]
Substituting the values of \( a \) and \( d \):
\[ a_n = 114 + (n - 1) \cdot 7 \]

### Step 3: Set up the inequality for the nth term
We want to find the largest \( n \) such that \( a_n < 245 \):
\[ 114 + (n - 1) \cdot 7 < 245 \]

### Step 4: Simplify the inequality
Subtract 114 from both sides:
\[ (n - 1) \cdot 7 < 245 - 114 \]
\[ (n - 1) \cdot 7 < 131 \]

### Step 5: Divide by the common difference
Now, divide both sides by 7:
\[ n - 1 < \frac{131}{7} \]
Calculating \( \frac{131}{7} \):
\[ \frac{131}{7} \approx 18.7142857 \]
Thus,
\[ n - 1 < 18.7142857 \]

### Step 6: Solve for \( n \)
Now, add 1 to both sides:
\[ n < 19.7142857 \]
Since \( n \) must be a whole number, we take the largest integer less than 19.7142857, which is 19.

### Conclusion
Therefore, the number of terms of the sequence below 245 is:
\[ \boxed{19} \]