To find how many terms exist in an arithmetic progression, we use the formula for the $n$-th term of an arithmetic progression:

$a_n = a_1 + (n - 1) \cdot d$

Where:

• $a_n$ is the $n$-th term,
• $a_1 = 7$ is the first term,
• $d = 5$ is the common difference,
• $n$ is the number of terms.

The progression stops before the number 1000, so the last term must satisfy:

$a_n < 1000$

Substituting the given values into the formula:

$a_n = 7 + (n - 1) \cdot 5$

Now, solving for $n$ when $a_n < 1000$:

$7 + (n - 1) \cdot 5 < 1000$

First, subtract 7 from both sides:

$(n - 1) \cdot 5 < 993$

Now, divide both sides by 5:

$n - 1 < 198.6$

Since $n$ must be an integer, we round 198.6 down to 198:

$n - 1 \leq 198$

Therefore, $n \leq 199$.

Thus, the number of terms in the arithmetic progression is 199.

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. What is the sum of all the terms in this arithmetic progression?
2. What is the 100th term of this arithmetic progression?
3. How would the result change if the common difference were negative?
4. Can you find the general formula for the sum of an arithmetic progression?
5. How many terms are there if the progression stops at exactly 1000?

Tip: In arithmetic progressions, always check if the sequence includes or excludes the final value before calculating the number of terms.