A monument made of a certain number of rows consisting of cube-shaped bricks starts with a row of 34 bricks. The row above has 31 bricks, the row above that has 28 bricks, and so on such that each row has three fewer bricks than the row below it. A portion of the monument is shown below. Later a student notices that the total number of bricks used in the monument is just enough to create the brick floor of a rectangular patio that is one layer of bricks, seven times as long as it is wide. (No bricks are broken to make the floor.) How many rows of bricks does the monument have?
The patio is to be rectangular.
If it's n bricks wide, it will be 7n bricks long, requiring 7n^2 bricks in total.
Treating the bricks in the monument as an AP, first term is 34 and common diff is -3.
If there are k rows of bricks, the number of bricks will be
\(\frac{k}{2}(2.34+(k-1)(-3))= \frac{k}{2}(71-3k)\)
For this to be of the form 7n^2, k = 7 (trial and error !), getting us 175 bricks.
175 = 5 * 35.
