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Mathematics 85 Online
OpenStudy (anonymous):

hello! can someone help explain "the nth term of an arithmetic sequence please?" I'm looking at my notes and it says We can use the following formula to find the nth term in any arithmetic sequence. tn=t1+(n-1)d For example, find the sixth term in the arithmetic sequence defined t1 = -2 and tn = tn-1 + -3. We know that the first term is -2, d is equal to -3, and n is equal to 6 since we are looking for the sixth term. t6=-2+(6-1)(-3) t6=-17 I don't understand where the 6 comes from.

OpenStudy (anonymous):

is it because it says "find the sixth term?"

OpenStudy (phi):

yes. you replace the n with 6

OpenStudy (anonymous):

oh okay! thank you :)

OpenStudy (phi):

to find the nth term... that means n is the number of the term

OpenStudy (phi):

notice that how this is written can be confusing we should write it \[ t_n=t_1+(n-1)d \] the subscript n on the t means the nth term the subscript 1 on the t means the first term if we want a particular term, replace n with that number, and then do the arithmetic (we also need to know d, the difference between 2 terms)

OpenStudy (anonymous):

okay that makes a lot more sense. :) thanks!

OpenStudy (anonymous):

is geometric similar to this?

OpenStudy (phi):

also , we have to "do some figuring" on \[ t_1 = -2 \\ t_n = t_{n-1} + -3. \] the first line says the first term is -2 the second line says the nth term is the previous term plus -3 we use that info to figure out that the difference between terms is -3 (i.e. d=3)

OpenStudy (phi):

arithmetic sequence have terms that have the same difference the simplest would be 1,2,3,4,... (the difference is 1) or 2,4,6,8,... (the difference is 2) in other words, we add the same number to each term to get the next term ARITHMETIC= ADD geometric sequence are terms multiplied by the same factor geometric = multiply

OpenStudy (phi):

here is a geometric (the factor is 2) 1,2,4,8,16,32,...

OpenStudy (anonymous):

ah :) okie

OpenStudy (anonymous):

one more thing, so there's this equation, and it's \[3=t _{1}+3d\] \[19=t _{1}+7d\] the answer (to that part) is 16=4d but it doesn't say to subtract, and when we do, 16 is -16.

OpenStudy (phi):

It looks like those equations came from the info: the 4th term of an arithmetic sequence is 3 the 8th term of this same arithmetic sequence is 19 and they want to find the first term, and the difference between the terms i.e. find t1 and d

OpenStudy (anonymous):

here's what the notes say. "Given that the 4th term of an arithmetic sequence is 3 and the 8th term is 19, find the 20th term." \[t _{4}=t _{1}+(4-1)d\] that equals \[3=t _{1}+3d \] and then there's an equation for the 8th term as well. I understand that part, what I don't understand is how they got 16=4d from the two equations

OpenStudy (phi):

it is well-known (to those who know it), that if you have n variables and n equations, you can solve for all n variables. in this case, we have 2 unknowns t1 and d and 2 equations. to solve a "system of equations" we can use elimination or substitution

OpenStudy (anonymous):

I think they said system of equations. hold on let me keep watching the thingy

OpenStudy (anonymous):

oh..

OpenStudy (anonymous):

I pressed "Play" and the guy said "subtract the first equation from the second equation."

OpenStudy (anonymous):

okay, subtracted the two equations, but the answers come out negative, in the notes, they're positive.

OpenStudy (phi):

if we use elimination, we often write the problem so all the unknowns are on the left side (we don't have to, but it helps to keep things organized) so let's write them as t1+ 3d= 3 t1 + 7d= 19 the idea with "elimination" is to get rid of one of the unknowns by multiplying one of the equations by a factor and then adding the two equations Here we see that t1 + -t1 would give us zero. (t1 "goes away" ) multiply the bottom equation by -1 t1 + 3d = 3 -t1 + -7d = -19 (each term gets multiplied by -1) --------------- 0 -4d = -16 0 - 4d is just -4d , so -4d = -16 divide both sides by -4 d = 4

OpenStudy (anonymous):

http://media.ccsdde.net/aventa/Algebra%202/ALG2x-HS-A09/b/unit10/a2_10.A.9.html i don't know if it'll allow you in, but this is what I'm looking at. I understand the elimination part, just that the "-4d=-16" is "4d=16"

OpenStudy (anonymous):

like I get that the answer is negative I just don't know why in the notes it's positive, the final answer is d=4 either way.

OpenStudy (phi):

you can do the subtraction in the other order t1 + 7d= 19 t1 + 3d = 3 <-- multiply this equation (both sides , all terms) by -1 t1 + 7d= 19 -t1 + -3d= -3 ------------ 0 + 4d = 16 4d= 16 d= 4 same answer.

OpenStudy (anonymous):

OHHH okay I think that's what they did.

OpenStudy (phi):

you say toe MA toe , I say TAH MA toe

OpenStudy (anonymous):

thank you oh so much xD

OpenStudy (phi):

I assume you know how to find t1. Pick one of the equations and solve for t1 t1 + 3d = 3 with d=4 t1 + 3*4 = 3 t1+12= 3 t1= 3-12 t1= -9 and finally, fn= -9 + (n-1)*4 now you can find any term

OpenStudy (phi):

*\[ t_n= -9 +4(n-1) \]

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