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.