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jillian.sala
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Linear Sequences

by jillian.sala Sun Apr 26, 2009 5:35 pm

I'm studying the MGMAT flash cards and am having trouble with linear sequences.

The question is, "What is the 25th term of this sequence?"

S(sub n) = S(sub n-1) -10 and S(sub 3) = 0

I don't understand how I find the direct sequence formula for this. Is this somewhere in the strategy guides?

Thanks!
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Re: Linear Sequences

by RonPurewal Sun May 03, 2009 4:56 am

jillian.sala Wrote:I'm studying the MGMAT flash cards and am having trouble with linear sequences.

The question is, "What is the 25th term of this sequence?"

S(sub n) = S(sub n-1) -10 and S(sub 3) = 0

I don't understand how I find the direct sequence formula for this. Is this somewhere in the strategy guides?

Thanks!


hey -

you don't need to find a direct formula.

in this problem, as in many other sequence problems, you should just LOOK FOR A PATTERN, and then extrapolate the pattern that you find.

with the data you've posted:
3rd term = 0
4th term = -10
5th term = -20
6th term = -30
etc.

it's pretty clear what's going on here.

the 25th term is twenty 10's below the 5th term:
25th term = 5th term - (20)(10) = -20 - 200
= -220

trying to find an explicit formula here is pretty much a waste of time.
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Re: Linear Sequences

by santhosh.ren Wed Dec 22, 2010 3:13 pm

The formula is -
S(n) = 20n + k, where k is a constant.
nth term = difference in terms * n + k, where k is a constant.

So, in this sequence, the first step is to find the difference by which the terms are increasing/decreasing. The difference here between terms is 20.

So, S(n) = 20n + k. To find k, substite one of the given terms in the sequence.
-2 = 20(1) + k (since -2 is the first term in the sequence, n=1)
k = -22.

So, the formula to find the nth number in this series would be:
S(n)=20n-22.
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Re: Linear Sequences

by jnelson0612 Sun Dec 26, 2010 10:19 am

santhosh, please go back and read Ron's explanation. Your formula is not going to yield the correct result.
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Re: Linear Sequences

by NanoBotZ44 Thu Sep 20, 2012 1:09 pm

As per the pattern:

S4 = S3 - 10
S5 = S4 - 10 = S3 -20
S6 = S5 - 10 = S3 - 30
S7 = S6 - 10 = S3 - 40 = S3 - 10x(7-3)

S(n) = S(3) - 10x(n-3)
S(25) = S(3) - 10x(25-3) = 0- 220 = -220
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Re: Linear Sequences

by jnelson0612 Sat Sep 22, 2012 9:02 pm

NanoBotZ44 Wrote:As per the pattern:

S4 = S3 - 10
S5 = S4 - 10 = S3 -20
S6 = S5 - 10 = S3 - 30
S7 = S6 - 10 = S3 - 40 = S3 - 10x(7-3)

S(n) = S(3) - 10x(n-3)
S(25) = S(3) - 10x(25-3) = 0- 220 = -220


Very nice!
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MattD517
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Re: Linear Sequences

by MattD517 Sat Jul 19, 2014 11:40 am

Just one quick question on this: My flash card (iphone app) adds the condition "for all integers n >= 2" in the question stem. Now I completely understand how we are getting to -220 as S25 without this condition. However, the condition that n >= 2 is throwing me off a bit. How should I interpret this?

Originally I was thinking that where n = 2, this would be the first in the sequence since n < 2 is implied to be invalid by this condition. Therefore I had originally calculated the 25th term as where n = 26 not where n = 25, and therefore I calculated the 25th term as -230.

Can anyone offer guidance on how to think about such a condition when presented with a sequence?

Thank you!
-Matt
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Re: Linear Sequences

by MattD517 Sat Jul 19, 2014 11:48 am

Hello, quick question on this one. My flash card for this question adds the condition "for all integers n >= 2" in the question stem. Now I completely understand how S25 = -220, but the condition that n >= 2 is throwing me off a bit.

My original answer to this question assumed that the 25th term of this sequence was actually S26, since S1 is "not valid" based on the condition that n >= 2. Therefore I answered -230 for the 25 term (S26, not S25).

In other words, I thought S2 was actually the first term. Can anyone offer guidance on how to interpret this condition? Any help appreciated.

Thank you!
-Matt
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Re: Linear Sequences

by tim Sun Jul 20, 2014 6:21 am

Just ignore the n>2 condition for this problem. You bring up a good point, and I think the question could certainly have been worded better. On the GMAT, unless they peg the nth term to S(n), they aren't going to ask for the nth term because it isn't defined. Instead, they will ask for S(n), which in this case is quite well defined. Go back to this question and ask yourself whether you would have gotten it right had the question asked for S(25) instead of the 25th term. If so, then you understand all you need to about this one!
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Re: Linear Sequences

by RonPurewal Wed Jul 23, 2014 3:16 am

Au contraire, there's actually a GMAT problem that does this EXACT thing: it gives a recursive (based on previous value) relationship, starting with n = 2.
This is absolutely necessary, because the relationship is nonsense if n = 1.
Look at it"”"”it contains both S(n) and S(n - 1). If n = 1, you'd need to have an S(0).
If you want to specify that this sequence starts with term no. 1, then you have to start the relationship at n = 2. The condition n ≥ 2 absolutely implies that the sequence starts with n = 1, since you need both S(1) and S(2) for that equation to mean anything when n = 2.
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Re: Linear Sequences

by RonPurewal Wed Jul 23, 2014 3:18 am

By the way"”"”We can actually generalize here:
ALL sequences in official problems start with the n = 1 term.
Every single one of them, in the problems published to date.

So, it's a safe bet that this will continue.
In other words, it's safe to just go ahead and assume that the first term is "term no. 1".

There are many areas of math, physics, and engineering in which sequences traditionally start with term number 0, but the GMAT has never, ever, ever used such a sequence.
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Re: Linear Sequences

by tim Mon Jul 28, 2014 1:36 pm

Thanks for clearing that up, Ron! It looks like based on known GMAT problems my extra note of clarification isn't altogether necessary. This is certainly good news for students who might otherwise worry more than they need to. :)

BTW Ron would you mind posting the problem number that you're referring to so students can see it in action on an actual GMAT problem?
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Re: Linear Sequences

by RonPurewal Wed Jul 30, 2014 9:59 am

tim Wrote:Thanks for clearing that up, Ron! It looks like based on known GMAT problems my extra note of clarification isn't altogether necessary. This is certainly good news for students who might otherwise worry more than they need to. :)

BTW Ron would you mind posting the problem number that you're referring to so students can see it in action on an actual GMAT problem?


Here is a thread on which, amusingly enough, this exact aspect of this exact problem is explained by none other than you:

post48299.html#p48299
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Re: Linear Sequences

by tim Thu Jul 31, 2014 10:08 am

Well, that problem doesn't refer to ordinal terms, which was the issue I noted on this one. But you bring up a good point about the fact that we can rely on the GMAT's consistency regarding this convention regardless of how the broader math community would interpret it. :)
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Re: Linear Sequences

by RonPurewal Sun Aug 03, 2014 5:30 pm

Yep. Always the same.