Question

Each of these sequence or series formulas involves four quantities. For each formula, describe the four quantities. Then explain how you can find the fourth quantity if you know the values of the other three.
a)An=A1+(n-1)d
b)Sn=n/2(A1+An)
c)An=A1r^n-1
d)Sn=(A1(1-r^n)/(1-r)

Answer 1

someone please help!

Answer 2

\[a_n=a_1+(n-1)d\]for example?

Answer 3

\(a_1\) is the first term, you can tell because of the little 1
\(n\) is the number of terms, and so \(a_n\) is the "nth" term
and \(d\) is the "common difference"
how is that for english?

Answer 4

i still dont get it

Answer 5

Don't get it?

If \(\large\rm a_{\color{orangered}{1}}\) is the \(\large\rm \color{orangered}{1st}\) term,
then is there some notation we can use to write our \(\large\rm \color{orangered}{4th}\) term?

Answer 6

Your "An=A1+(n-1)d" was one of the first formulas for sequences that I had to learn. I learned it using slightly different notation: n is the number of terms in an arithmetic sequence; d is the common difference (the jump from one term of the sequence to the next); a is the first term; l is the last term. Invent some sequence (e. g., a=3, n=5, d=2) and use this formula to predict the last term. Write out the sequence, term by term.

Answer 7

@shaeelynn Are the symbols the most confusing part for you?

Answer 8

yes i don't really understand it my teacher gave us this assignment and didn't teach us how to do ittt

Answer 9

Well, then, we should probably start with the basics.

Answer 10

You've probably said this sequence before in a math class:
2, 4, 6, 8, 10, 12, 14, 16, ...

Answer 11

A "sequence" is just a list of numbers that follows a pattern.

Answer 12

The sequence we said above, is just the "counting by twos" sequence.

Answer 13

Make sense so far?

Answer 14

yess

Answer 15

Cool, now mathematicians wanted a way to write down this sequence using a formula. It gets a bit exhausting to have to keep writing a bunch of numbers just to see the pattern.

Answer 16

What they thought is something like this.
Step 1) We'll give each number a name. We'll call them "a_n" that's a subscript n.

Answer 17

For example,
a1 = 2 it is the 1st term
a2 = 4 it was the 2nd term
a3 = 6 it was the 3rd term
...
etc.

Answer 18

Any questions on that part?

Answer 19

We are basically just using "a_1" as a name for the first number of the sequence (also called the first "term"). "a_2" is the name for the second number of the sequence... and so on. Make sense?

Answer 20

yes that makes sense

Answer 21

OK, so just so you can see how this works. I (being a fancy mathematician) might ask, "Well, what is a_5?" What I am really asking is what is the fifth term of the sequence.

Answer 22

Well, of course, we could go back to our list of numbers that I posted earlier, and we could see that a_5 = 10. The fifth number of the sequence was 10.

Answer 23

Alright, so at this point, you should be able to understand what a_1, a_2, a_3, and so on mean.

Answer 24

Now for the next step.

Answer 25

Instead of writing out all the numbers, I could be very clever and write this:
a_n = 2*n

Answer 26

Ack! What does this mean?

Answer 27

Well, this is the formula for the "counting by twos" sequence.

Answer 28

Here's how it works. Let's find the first number of the sequence. We choose n = 1, the first.

Answer 29

Plug in n = 1, and we magically get:
a_n = 2*n
a_1 = 2*1 -> a_1 = 2

Answer 30

And that matches what we got before! Try it with n = 2. The second term.

Answer 31

a_n = 2*n
a_2 = 2*2 -> a_2 = 4

Answer 32

It keeps working! Now, your turn. What is a_3, using the formula?

Answer 33

If you need a hint let me know.

Answer 34

6?

Answer 35

Yes, that is correct!

Answer 36

Sorry, my internet was lagging.

Answer 37

What's really nice about the formula is it let's you figure out things that would be impossible! Like what is the 200th term of the sequence? With the formula it's easy:

a_n = 2*n
a_200 = 2*200
a_200 = 400

Answer 38

OK, that is the basics for sequences. We have to make things a little harder in order to answer your original question.

Answer 39

With our formula method, we can invent new sequences that we haven't seen before:
Let's try one that is very similar to (a) of your question.

Answer 40

a_n = a_1 + (n-1)*3

Answer 41

This formula looks really complicated! But, it's basically the same idea.

Answer 42

This formula needs us to tell it what a_1 is, so let's just say that a_1 = 1.

Answer 43

Now, we can figure out what the second term has to be. a_2
Just plug in n = 2.
a_n = 1 + (n-1)*3
a_2 = 1 + (2-1)*3 -> a_2 = 1 + (1)*3 -> a_2 = 1+3 -> a_2 = 4

Answer 44

See how I did that?

Answer 45

yes i do

Answer 46

Great, if you keep plugging in different n's like n = 3, n = 4, n = 5, you will get more and more terms of the sequence. For the formula I made up, this would be
a_1 = 1,
a_2 = 4,
a_3 = 7,
a_4 = 10,
a_5 = 13.
and so on.

Answer 47

Now, what I'd recommend is rereading the chapter and following the examples in the book. I could try and go through all the different ways to set up and solve these problems but it would get very messy. Before I stop though, let's revisit satellite's answer to understand what he was saying.

Answer 48

Part a) a_n = a_1 + (n-1)*d

The "n"th term is found by taking the "first term" and adding "(n-1)" multiplied by the difference "d."

It turns out that the number to the right of the (n-1) [the d] predicts the difference between the terms. You can see this in my example above (d was 3) and the numbers jump by 3.

Answer 49

Your book should teach you how to solve the problems. You might also want try searching for "sequences and series" on Khan Academy.

Answer 50

Oh, this might be helpful, here is a quick translation of the common symbols used in this part of math:

a_n = the "nth term of the sequence"
a_1 = the "1st term of the sequence"
d = the "common difference"
r = the "common ratio"

Answer 51

If you don't like your textbook, this series of videos might be helpful: https://www.khanacademy.org/math/precalculus/seq-induction

Answer 52

thank youu

Answer 53

you're welcome