# Thread: Sequence and Series Question

1. ## Sequence and Series Question

Hey everyone. I haven't been on in quite some time, but I guess that was due to not having school, although I do think I should've helped out those who had similar problems as I did, but I am sorry. Anyways, I just started Precalculus, and I had a question that I was quite confused on. Well anyway, here goes....

How many terms are in the sequence 4,7,10.....40?

Now i know I could do this the long way by doing it manually, but I forget the formula or way to do it to get it in a much simpler way. I would appreciate anyone that can help me with this and bid you all a good day. Edit: I also forgot that I put this in the wrong section with elementary/middle school math questions, when it belongs in the high school math section, namely in Pre-Calculus, so I am sorry about that.

2. Originally Posted by Norman Smith Hey everyone. I haven't been on in quite some time, but I guess that was due to not having school, although I do think I should've helped out those who had similar problems as I did, but I am sorry. Anyways, I just started Precalculus, and I had a question that I was quite confused on. Well anyway, here goes....

How many terms are in the sequence 4,7,10.....40?

Now i know I could do this the long way by doing it manually, but I forget the formula or way to do it to get it in a much simpler way. I would appreciate anyone that can help me with this and bid you all a good day. Edit: I also forgot that I put this in the wrong section with elementary/middle school math questions, when it belongs in the high school math section, namely in Pre-Calculus, so I am sorry about that.
These are incrementing by 3 at each step, and the n-th term is 1+3n.

If N is the number of terms then:

1+3N=40,

so:

N=39/3=13.

RonL

3. How many terms are in the sequence 4,7,10.....40?

Let's see how the sequence goes.
1st term = 4
2nd term = 7
2nd term minus first term, 7-4 = 3

3rd term is 10
3rd term minus 2nd term, 10 -7 = 3

Umm, the sequence is Arithmetic. The common difference between terms is 3.

a2 = a1 +d
a3 = a2 +d = a1 +d +d = a1 +2d = a1 +(3-1)d
....
an = a1 +(n-1)d

So, since a1 = 4 and an = 40, and d=3, then,
40 = 4 +(n -1)(3)
(n-1) = (40 -4)/3
n -1 = 12
n = 12 +1 = 13

Therefore, there are 13 terms in that sequence.

4. Well thank you both for the help, and I sincerely appreciate it, and also understand it, however I am having difficulty with another problem.

It states: Cylindrical tiles of uniform size are stacked in the form of a triangle. There are 21 tiles on the bottom row. Find the total number of tiles in the stack.

5. Post new questions in a new thread.

Stack something.

1 gives 1
Put that 1 on two others. That's three.
Make a base 3 wide and move the first three on top. That's six.

Now what?

6. Originally Posted by Norman Smith Cylindrical tiles of uniform size are stacked in the form of a triangle. There are 21 tiles on the bottom row. Find the total number of tiles in the stack.
This sequence is the same as the first one.
Here,
a1 = 21
an = 1
d = -1

That is if the tiles are stacked up in such as way that a tile is resting on two tiles below it. That is, the center of the tile above is aligned to the meeting of the two tiles below.
a1 = 21
a2 = 20
a3 = 19
.
.
an = 1

So, again,
an = a1 +(n-1)d
1 = 21 +(n-1)(-1)
1 = 21 -n +1
n = 21 +1 -1
n = 21

Then, the sum of an Aritmetic series is
Sn = (n/2)(a1 +an)
So,
S21 = (21/2)(21 +1)
S21 = 21(11)
S21 = 231

Therefore, there are 231 tiles in the stack. ------answer.

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