Yes, you can definitely use that method to find the sum of the first 15 terms. This formula is a simpler version of the sum formula for an arithmetic sequence, particularly when you already know the first term ($U_1$) and the 15th term ($U_{15}$).

The formula is: $S_{15} = \frac{n}{2} \times (U_1 + U_{15})$ where $n$ is the number of terms.

Let's apply this to your sequence:

1. First term ($U_1$) = 3
2. 15th term ($U_{15}$): We need to calculate this first using the general term formula: $U_{15} = a_1 + (15-1) \times d = 3 + 14 \times 4 = 3 + 56 = 59$
3. Number of terms ($n$) = 15

Now, substitute these values into the sum formula: $S_{15} = \frac{15}{2} \times (3 + 59) = \frac{15}{2} \times 62 = 15 \times 31 = 465$

So, the sum of the first 15 terms is indeed 465, using your suggested method.

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Would you like to go over anything else?

Related Questions:

1. How would you find the 15th term directly using the sequence formula?
2. What is the advantage of using the formula $S_n = \frac{n}{2} \times (U_1 + U_n)$?
3. How can you find the sum if you only know the first term and the common difference?
4. Can this method be applied to any arithmetic sequence, regardless of the number of terms?
5. How does the sum formula for an arithmetic sequence differ from that of a geometric sequence?

Tip:

Using the formula $S_n = \frac{n}{2} \times (U_1 + U_n)$ is particularly useful when you have both the first and the last term readily available.