To find the 65th term of the arithmetic sequence $-10, 9, 28, \dots$, we follow these steps:

Step 1: Identify the first term and common difference

• The first term $a_1$ is $-10$.
• The second term $a_2$ is $9$.
• The third term $a_3$ is $28$.

The common difference $d$ is: $d = a_2 - a_1 = 9 - (-10) = 19.$

Step 2: Use the formula for the nth term of an arithmetic sequence

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

For the 65th term ($n = 65$): $a_{65} = -10 + (65 - 1) \cdot 19.$

Simplifying: $a_{65} = -10 + 64 \cdot 19.$

First, calculate: $64 \times 19 = 1216.$

Now: $a_{65} = -10 + 1216 = 1206.$

Final Answer

The 65th term of the sequence is 1206.

Would you like further details or additional examples?
Here are related questions for more practice:

1. What is the 50th term of the sequence $2, 5, 8, \dots$?
2. Find the sum of the first 65 terms of the sequence $-10, 9, 28, \dots$.
3. What is the common difference of the sequence $-7, -3, 1, \dots$?
4. If $a_1 = 4$ and $d = 6$, what is the 20th term of the sequence?
5. Determine the arithmetic mean of $-10$ and $28$.

Tip: In arithmetic sequences, the common difference stays constant between consecutive terms.