Question

I need someone to explain how to do this.. How do you write an equation for the nth term of the Arithmetic Sequence 4, 1, -2, -5? I know the difference is -3 and a1 is 4 (i think) but what do i use that for and how do i write it and what is it?

Answer 1

the formula to use is \[a_n = a_1 + (n-1)d\] where \(n\) is the number of the term you seek, \(d\) is the common difference, and \(a_1\) is the value of the first term.

Answer 2

so would the equation just be An= 4 + (n-1) (-3)

Answer 3

Let's find out!

\[a_2 = a_1 + (2-1)(-3) = 4 + 1(-3) = 4-3 = 1\]\[a_3 = a_1+(3-1)(-3) = 4 +2(-3) = 4-6 = -2\]\[a_4 = 4+3(-3) = 4-9 = -5\]

Does that look like it produces the right results?

Answer 4

Oh! so i don't need anything else? i thought there was more to it!

Answer 5

it came out with the right answers when i did it too!

Answer 6

That's all there is to that problem, but there are some similar problems that we could do so you know how to do them as well.

For example, what if you know the 1st and 3rd terms only? Could you figure out the formula?

Answer 7

i dont know how you'd do that..

Answer 8

Well, let's try it with a new sequence.

1, ???, 5, ???, etc.

First, what's the value of \(a_1\)?

Answer 9

1

Answer 10

Good. So, what's the formula going to be (just use letters where you don't know the values)?

Answer 11

\[An _{?} = 1 (n-1) (d)\]

Answer 12

1+ (n-1) (d)*

Answer 13

very good!
by the way, to do a subscript in the equation editor, you write _ before the thing you want as a subscript. so a_n = a_1 + (n-1)d is what I type in. If you need a subscript to have more than one digit, you wrap it in { }, so \(a_{23}\) is produced with a_{23}

Answer 14

So we have \[a_n = 1+(n-1)d\]Now the third term of the sequence is 5. Can we use that information to figure out the value of \(d\)?

Answer 15

yes but i dont know how..

Answer 16

Well, what is the formal name for the 3rd term?

Answer 17

if theres an equation ment to find just the 3rd term my teacher didnt show it to us..

Answer 18

oh, you know it, you figured it out in the other problem :-)

\(a_3\) is the third term, isn't it? The sequence goes \(a_1, a_2, a_3, a_4, ..., a_n\)

Answer 19

So we can write \[a_3 = a_1 + (n-1)d\]but if we are looking at \(a_n = a_3\) then we have \(n = 3\), right? And we already know that \(a_1 = 1\) so that gives us\[a_3 = 1 + (3-1)d\]
Now can you find the value of \(d\)?

Answer 20

oh oh! yeah i knew that.. and so a_3 = 1+ (4) d

Answer 21

no, not quite...

Answer 22

wait im very confused

Answer 23

\[a_3 = 1 +(3-1)d\]\[a_3 = 1 + 2d\]\[5 = 1 + 2d\]right?

Answer 24

I think you added instead of subtracting...

Answer 25

you were excited :-)

Answer 26

oh im sorry im not thinking straight and i did..

Answer 27

Okay, so \(d = \)

Answer 28

d is 2?

Answer 29

so the equation would be A1= 1 + (n-1) (2)?

Answer 30

Yes!

So our sequence is \[a_n = 1 + (n-1)2\]or more usually \[a_n = 1+2(n-1)\](writing it the other way on OpenStudy will make people think that maybe you copied and pasted from something like \((n-1)^2\) and ended up with \((n-1)2\) which wasn't true)

Answer 31

Okay, now for the final variation :-)

What if I told you that you needed to find the formula for this sequence? ???, 4, ???, ???, 13

Answer 32

2nd and 5th place.. hold on ill do it

Answer 33

I've got a nice, shiny medal ready to award to you when you get it :-)

Answer 34

do i need to find all of the empty spaces or just one?

Answer 35

find the formula, and I'll take it for granted that you could find any space I asked for...

Answer 36

i am stuck already..

Answer 37

well, write out what you can, and I'll help you along.

Answer 38

so you start with An= a1 + (n-1) (d)

Answer 39

yes.

Answer 40

but this time, you don't know the value of \(a_1\), do you?

Answer 41

no..

Answer 42

doesn't matter :-)

Answer 43

write out what you do know for the two terms you know.

Answer 44

a2= a1 + (2-1) (d)

Answer 45

it's going to turn out to be essentially the same problem as finding the y-intercept of a line given two points on the line

Answer 46

wait is that part right?

Answer 47

yes, so far so good...

Answer 48

now what?

Answer 49

do the same with the other term you know...

Answer 50

ok i did

Answer 51

so what are the two equations you have, with all the numbers you know substituted in?

Answer 52

the one i said already and a5= a1 + (5-1)(d)

Answer 53

okay, but you haven't substituted in ALL of the numbers you know...

Answer 54

what others do i know??

Answer 55

don't you know the actual values of \(a_2\) and \(a_5\)?

Answer 56

oh yeah do i plug that in where a1 is?

Answer 57

No, you plug it in where its name is...

if \(a_2 = 4\), then the equation is \[4 = a_1+(2-1)d\]\[4=a_1 + d\]

Do the same for \(a_5\), please...

Answer 58

13= a1+ (4) (d)

Answer 59

So we have two equations:
\[4 = a_1 +d\]\[13 = a_1 + 4d\]

Is it possible to find the values of \(a_1\) and \(d\) from that alone?

Answer 60

i mean i bet it is but i dont know how..

Answer 61

Haven't you done systems of equations yet? Substitution, or elimination?

Answer 62

not really no..

Answer 63

Hmm, I thought that usually happened before arithmetic sequences, but maybe not.

Okay, well, can you solve \[4 = a_1 +d\]for \(a_1\)? You won't get a number, you'll get an equation, but with only \(a_1\) on one side of the = sign.

Answer 64

\[4 = a_1 + d\]Subtract \(d\) from both sides:\[4-d = a_1 + d - d\]collect like terms:\[4-d = a_1 \cancel{+d}\cancel{-d}\]\[a_1 = 4-d\]right?

Answer 65

yes..

Answer 66

Okay. Now take the second equation, and replace \(a_1\) with \(4-d\). What do you get?

Answer 67

i was doing it wrong i got it now hold on!

Answer 68

i am messing it up i think..

Answer 69

which do i do first? subtract the 4 or divide the other 4?

Answer 70

\[13 = a_1+4d\]and we found that \[a_1 = 4-d\]

\[13 = (4-d) + 4d\]
Can you solve that for \(d\)?

Answer 71

divide the four first right

Answer 72

No, get rid of the parentheses first.

Answer 73

that gives us \[13 = 4-d+4d\]
what is \[-d + 4d\]?
Hint: \[-d = -1d\]

Answer 74

9= -d+4d

Answer 75

i subtracted the 4 from 13

Answer 76

Good so far, keep going

Answer 77

do i divide by 4?

Answer 78

No. what does \(-d + 4d=\)

Answer 79

9

Answer 80

no, just simplify that expression...

Answer 81

oh... wait you can't i didnt think

Answer 82

sure you can. it's just the same as \[4d - d\] or \[4d - 1d\]

Answer 83

oh!

Answer 84

is there a big dent in your forehead now? :-)

Answer 85

yeahh..

Answer 86

so \[9 = -d+4d\]\[d = \]

Answer 87

i am still confused..

Answer 88

so divide the four now??

Answer 89

or add d to 13??

Answer 90

oh, don't be silly :-)
\[9 = -d + 4d\]\[9 = 3d\]\[d=\]

Answer 91

oh

Answer 92

i got it...

Answer 93

\[d = \]
prove that you got it :-)

Answer 94

3

Answer 95

right. so now we know \(d = 3\), what is \(a_1=\)

Answer 96

1,4,7,10,13

Answer 97

an= 1+ (n-1)(3)

Answer 98

right!

so the process we used to solve that system of two equations in two unknowns is called substitution. we take one of the equations, solve it for one variable in terms of the other, and then substitute it into the second equation, giving us an equation in terms of just one variable. We solve that, and then use the first equation to find the value of the variable we substituted.