To find the $n$th term ($a_n$) of an arithmetic sequence, we use the formula:

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

Where:

• $a_1$ is the first term,
• $d$ is the common difference,
• $n$ is the term number.

Let's apply this formula to each of the given sequences:

1. $a_1 = 7$, $d = 2$, $n = 16$

$a_{16} = 7 + (16-1) \cdot 2 = 7 + 15 \cdot 2 = 7 + 30 = 37$

2. $a_1 = 5$, $d = -3$, $n = 15$

$a_{15} = 5 + (15-1) \cdot (-3) = 5 + 14 \cdot (-3) = 5 - 42 = -37$

3. $a_1 = x$, $d = 2x$, $n = 18$

$a_{18} = x + (18-1) \cdot 2x = x + 17 \cdot 2x = x + 34x = 35x$

Summary

1. $a_{16} = 37$
2. $a_{15} = -37$
3. $a_{18} = 35x$

Would you like details on how this formula works or any further explanations?

Here are five related questions:

1. What is the sum of the first 16 terms of the first sequence?
2. How would the sequence change if the common difference in the second sequence were positive?
3. Can you express the general term $a_n$ for the third sequence?
4. What happens to the nth term when $n$ is very large in an arithmetic sequence with a negative difference?
5. How would the nth term formula change if $n$ starts from a different index?

Tip: In arithmetic sequences, the common difference ($d$) remains constant, which is why the sequence grows or decreases linearly.