Find the nth Term Formula for Sequence: 3,6,9,12,15

Given the series whose difference between two jumped numbers is constant:

$3,6,9,12,15$

Describe the property using the variable $n$

We'll solve the problem by following these steps:

• Identify the first term ($a_1$) and the common difference ($d$).
• Use the arithmetic sequence formula $a(n) = a_1 + (n - 1) \times d$.
• Substitute the known values into the formula.

Now, let's go through each step:

Step 1: Identify the given information:
The first term of the sequence ($a_1$) is 3, and the common difference ($d$) is 3, as the difference between any two consecutive terms is constant and equal to 3.

Step 2: Use the formula for the $n$-th term of an arithmetic sequence:
$a(n) = a_1 + (n - 1) \times d$.

Step 3: Plug in the known values:
- First term $a_1 = 3$.
- Common difference $d = 3$.
Therefore, $a(n) = 3 + (n - 1) \times 3$.

Check the formula by substituting values:
- For $n = 1$: $a(1) = 3 + (1-1) \times 3 = 3$
- For $n = 2$: $a(2) = 3 + (2-1) \times 3 = 6$
- Continue checking for other values.

Since the formula correctly generates the sequence values, the description of the series is $a(n) = 3 + (n - 1) \times 3$.

Therefore, the correct answer is choice 3.