Question

What is the 35th term of the arithmetic sequence where a1 = 13 and a17 = -35 ?

-89
-86
-83
-80

Answer 1

Do you know the formula for computing an arbitrary term in an arithmetic sequence?

Answer 2

Is it an=a1+(n+1)d

Answer 3

Very close...

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

As an aid to remembering how the formula is written, remember that \[a_1 = a_1\]so\[a_n = a_1+(n-1)d\]and when \(n=1\) then we have \[a_1 = a_1 + (n-1)d\]\[a_1 = a_1 + (0)d\]\[a_1=a_1\]

If you had a plus sign in the \((n-1)\), to make that part equal 0 we would have to be talking about \(n=-1\) but we don't do that...

Answer 4

Okay, so we know that \[a_n = a_1 + (n-1)d\]\[a_1 = 13\]\[a_{17} = -35\]
Can you use that information to find \(d\)?

Answer 5

I've calculated it to be three, but this is where I believe I have already messed up

Answer 6

well, let's check:

\[a_n = a_1 + (n-1)(3)\]\[a_{17} = -35 = 13 + (17-1)3\]\[-35 = 13+(16)(3)\]Yeah, that isn't going to work out. But we can set it up the other way and work back to \(d\):

\[-35 = 13 + (17-1)d\]
What do you get if you solve that for \(d\)?

Answer 7

-3?

Answer 8

Very good. \[-35 -13 = 13-13 + 16d\]\[-48 = 16d\]\[-3=d\]

So what is our formula for \(a_n\) in this sequence, and what will it give us for the value of \(a_{35}=\) if we ask nicely? :-)

Answer 9

Would it be a35 = 13 + (17-1)(-3)?

Answer 10

write out the general formula for \(a_n\) first...

Answer 11

an =a1 + (n-1) d?

Answer 12

right, now plug in the known values for \(a_1\) and \(d\)

Answer 13

Alright

Answer 14

an = 13 + (n-1)(-3)

Answer 15

Yes. Now find the value of \(a_{35}\) with that formula. Hint: \(n\ne 17\)

Answer 16

I got -6.33333... I know this can't be right :c

Answer 17

just put the numbers in the formula and write it out for me...

Answer 18

a35 = 13 + (n-1) (-3)

Answer 19

I'm glad you recognize that that cannot be the right answer — way too many students lack that sense!

what is the value of \(n\)?

Answer 20

Thank you c: Is n 35?

Answer 21

yes. it's the same number as the subscript of \(a_n\)...

(I agree that is a somewhat confusing point at first!)

Answer 22

So I solve a35 = 13 + (35 - 1) (-3)?

Answer 23

@whpalmer4? :)

Answer 24

that would be the one...

Answer 25

-89? :D

Answer 26

you got it!

Answer 27

Thank you for your time and help! I appreciate it very much :)

Answer 28

so here's a related problem, of a sort I'm confident you'll get sooner or later

Answer 29

might as well learn how to do it now, right? :-)

Answer 30

sure!

Answer 31

say you have a sequence where you know \[a_3=5\]and\[a_7 = -3\]Can you find the value of \(a_{19}\)?

Answer 32

Would I solve like here? Or sorta similar at least, this problems looks kind of like the one we just solved. :)

Answer 33

well, the wrinkle is that you don't know the value of \(a_1\)...

Answer 34

Oh alright, I haven't learned this yet in class. I'd be happy to watch how to solve though :)

Answer 35

but that turns out not to matter, so long as you know two different values of the sequence

Answer 36

Okay

Answer 37

why don't you write out the equations (with the numbers plugged in) for both values of the sequence: \(a_3 = 5\) and \(a_7 = -3\)
that may give you a clue as to how to solve it...

Answer 38

Well, I don't really have time to solve it myself, I'm going to a family get-together here soon, but I can watch if we do it quickly

Answer 39

I'd be happy to learn :)

Answer 40

oh, you'll be surprised at how quickly you can do it. just write out the formula, just like you did for the previous problem. I'll do one, you do one.

\[a_3 = 5 = a_1 + (5-1)d\]
\[5 = a_1 + 4d\]

Answer 41

can you do the one for \(a_7 = -3\)?

Answer 42

\[a_7 = -3 = a_1 + (7-1)d\]\[-3=a_1 + 6d\]

Now we have a system of two equations in two unknowns:

\[5 = a_1 + 4d\]\[-3 = a_1 + 6d\]

You've solved things like that before, right?

Answer 43

Burp, brain fart!

\[a_3 = 5 = a_1 + (3-1)d\]\[5 = a_1 + 2d\]\[a_7 = -3 = a_1 + (7-1)d\]\[-3 = a_1+6d\]

now that's a system of two equations in two unknowns:

\[5 = a_1 + 2d\]\[-3 = a_1 + 6d\]multiply first equation by -1 and add them together:
\[-1*5 + -3 = -1*a_1 + a_1 + -1*2d + 6d\]\[-8 = 4d\]\[d=-2\]

Now we plug \(d = -2\) into either of the equations to find \(a_1\):

\[a_3 = a_1 + (n-1)(-2)\]\[5 = a_1 + 2(-2)\]\[5 = a_1 -4\]\[a_1 = 9\]

So our sequence is \[a_n = 9 + (n-1)(-2)\]

and \[a_{19} = 9 + (19-1)(-2) = 9 -36 = -27\]

Answer 44

Here's the first 19 terms of that sequence, so you can check my work :-)

\[\begin{array}{cc}
n & a_n \\ \hline \\1 & 9 \\
2 & 7 \\
3 & 5 \\
4 & 3 \\
5 & 1 \\
6 & -1 \\
7 & -3 \\
8 & -5 \\
9 & -7 \\
10 & -9 \\
11 & -11 \\
12 & -13 \\
13 & -15 \\
14 & -17 \\
15 & -19 \\
16 & -21 \\
17 & -23 \\
18 & -25 \\
19 & -27 \\
\end{array}\]