The image shows a problem that asks to find the number of terms in two arithmetic sequences. Let's break down each sequence and solve it step by step.

Sequence 1: $41, 37, 33, \dots, -79$

This is an arithmetic sequence where the first term $a_1 = 41$ and the common difference $d$ can be found by subtracting the second term from the first: $d = 37 - 41 = -4$ We are asked to find the number of terms $n$ in the sequence where the last term $a_n = -79$.

The formula for the $n$th term of an arithmetic sequence is: $a_n = a_1 + (n-1) \cdot d$ Substituting the known values: $-79 = 41 + (n-1) \cdot (-4)$ Simplify and solve for $n$: $-79 = 41 - 4(n-1)$ $-79 = 41 - 4n + 4$ $-79 = 45 - 4n$ $-124 = -4n$ $n = \frac{124}{4} = 31$ So, the number of terms in the first sequence is $\boxed{31}$.

Sequence 2: $9, 15, 21, \dots, 219$

This is another arithmetic sequence where the first term $a_1 = 9$ and the common difference $d$ is: $d = 15 - 9 = 6$ We are asked to find the number of terms $n$ where the last term $a_n = 219$.

Again, use the formula for the $n$th term: $219 = 9 + (n-1) \cdot 6$ Simplify and solve for $n$: $219 = 9 + 6(n-1)$ $219 = 9 + 6n - 6$ $219 = 3 + 6n$ $216 = 6n$ $n = \frac{216}{6} = 36$ So, the number of terms in the second sequence is $\boxed{36}$.

Summary:

• The number of terms in the first sequence is $31$.
• The number of terms in the second sequence is $36$.

Would you like further details or have any questions?

Here are five related questions to expand your understanding:

1. How would the calculation change if the common difference were positive?
2. How do you derive the general formula for the $n$th term of an arithmetic sequence?
3. What would happen if the last term of the sequence was not explicitly provided?
4. Can you find the sum of all terms in each sequence?
5. How would you approach finding the number of terms in a geometric sequence?

Tip: When solving arithmetic sequence problems, always ensure you correctly identify the common difference and apply the formula consistently to avoid errors.