# a. One row contains 17 bricks. Which row is this? b. How many rows of bricks are there? Explain your assumptions. A pile of bricks is arranged in rows. The number of bricks in each row...

a. One row contains 17 bricks. Which row is this?

b. How many rows of bricks are there? Explain your assumptions.

A pile of bricks is arranged in rows. The number of bricks in each row forms the arithmetic sequence 65,59,53.

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If we are calling the first row of bricks (the one with 65 bricks in it) Row 1, then I would say that it is Row 9 that has 17 blocks. I would also say that there are 11 rows total. Here are my assumptions (how I arrived at this result):

First of all, this is an arithmetic progression. That means that there will be the same difference between any two pairs of consecutive numbers. From the data you have given us, that difference is 6.

I am assuming that each row, therefore, will have 6 fewer blocks than the one before it. We start out with 65 and need to get to 17. 17 is 48 less than 65 and 6 goes into 48 8 times. That means that the row with 17 is 8 rows after the row with 65.

The bricks arranged in rows as follows:

65, 59, 53 , ..., 17, ....

We notice that each term of the sequence is 6 bricks less than the row before it.

Then to complete the sequence :

65, 59, 53, 47, 41, 35, 29, 23, 17, 11,5

We notice that the row which contains 17 brick is number 9 in the sequence.

a. Then the answer is the 9th row.

b. There are 11 rows

The given arithmetic sequence is:

65, 59, 53 ...

This can be expressed as:

(6*11 -1), (6*10 -1), (6*9 -1), ... (6*5 -1),(6*4 -1), (6*3 -1)

Pleas not that in the above series the last number is 17 which is assume to be the number of bricks in last row in pile of bricks. This give total of 9 term or rows in the series.

The above series can be further rewritten as:

(6*8) + (6*3 -1), (6*8) + (6*3 -1), (6*7) + (6*3 -1), ... (6*2) + (6*3 -1), (6) + (6*3 -1), (0) + (6*3 -1)

= (6*8) + 17, (6*7) + 17, (6*6) + 17, ... (6*2) + 17, 6 + 17, 17

Thus addition of all the eleven terms in the series is equal given by:

= (6*8) + (6*7) + (6*6), ... + (6*2), 6, + 0

+ 17 + 17 + 17 + 17 + 17 + 17

= [6 (8 +7 + 6 + ... + 3 +2 + 1] + (9*17)

= 6*36 + 153 = 216 + 153 = 369

Answer:

When the last row contains 17 bricks, there will be 9 rows.

Total number of bricks = 369

The first row contains R1 bricks = 65 bricks.

The second Row contains R2 bricks = 59 bricks.

The 3rd contains R3 = 53 bricks

You can see that R1-R2 =65 -69 = 6 = R3-R2 = 59-53 = 6. So the common reduction between the successive rows is 6 bricks ( and it is reduction).

S0 R 1 -Rn = 65 -17 = 48. It requires 48 /6 = 8 rows further from 65 to reach the R1 which is R9 or the 9 th row. Or you can visualise from the following :

R1 = 65,

R2 = 59,

R3= 53

R4 = 47,

R5 = 41

R6 = 35,

R7 = 29,

R8 = 23

**R9 = 17, the 9 th row.**