A pile of bricks has 85 bricks in the bottom row, 81 bricks in the second row, 77 bricks in the third row, and so on. There is only one brick in the top row. a) How many bricks are there in the 13th row? B) How many rows are there in the pile? In arethmetic series

This problem can be modeled using an arithmetic sequence. In an arithmetic sequence, each term after the first is found by adding a constant difference, \(d\), to the previous term. Let's solve this step by step.

### Part A: How many bricks are there in the 13th row?

The total number of bricks decreases by 4 bricks in each subsequent row (as \(85, 81, 77, \ldots\)):

- The first term (\(a_1\)) of this arithmetic sequence is 85.
- The common difference (\(d\)) is \(-4\).

The formula for the \(n\)-th term of an arithmetic sequence is:

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

For the 13th row, \(n = 13\):

\[
a_{13} = 85 + (13 - 1) \cdot (-4)
\]
\[
a_{13} = 85 + 12 \cdot (-4)
\]
\[
a_{13} = 85 - 48
\]
\[
a_{13} = 37
\]

Thus, there are **37 bricks in the 13th row**.

### Part B: How many rows are there in the pile?

In this case, we are looking for the total number of rows, where the last row has only 1 brick. We need to find \(n\) such that the number of bricks in the \(n\)-th row is 1. Using the arithmetic sequence formula again, we set \(a_n = 1\) and solve for \(n\):

\[
1 = 85 + (n - 1) \cdot (-4)
\]
\[
1 = 85 - 4(n - 1)
\]
\[
1 = 85 - 4n + 4
\]
\[
1 = 89 - 4n
\]
\[
4n = 89 - 1
\]
\[
4n = 88
\]
\[
n = \frac{88}{4} = 22
\]

Thus, there are **22 rows in the pile**.

### Final Answer:
- A) There are **37 bricks** in the 13th row.
- B) The pile has **22 rows**.

Would you like more details on any part of the solution, or do you have any questions?

Here are 5 questions to deepen your understanding:
1. How would the total number of bricks change if the common difference was \(-5\) instead of \(-4\)?
2. How would you find the total number of bricks in the entire pile?
3. Can you express the total number of bricks in the first 10 rows?
4. What is the general term for any arithmetic sequence?
5. How would you solve for \(n\) if the common difference was a fraction?

**Tip:** When working with arithmetic sequences, always verify the common difference and the first term to ensure accurate results.

A pile of bricks has 85 bricks in the bottom row, 81 bricks in the second row, 77 bricks in the third row, and so on. There is only one brick in the top row. a) How many bricks are there in the 13th row? b) How many rows are there in the pile?

Learn how to solve an arithmetic sequence problem involving a pile of bricks where each row contains fewer bricks than the row below. This detailed solution includes finding the number of bricks in the 13th row and determining the total number of rows in the pile. Suitable for students in Grades 8-10.

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