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In an infinite sequence of integers a1, a2, a3, , a1 = -30

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In an infinite sequence of integers a1, a2, a3, , a1 = -30 [#permalink] New post 23 Jun 2007, 13:53
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In an infinite sequence of integers a1, a2, a3,… , a1 = -30 and for all n > 1, an-1 + 4 <an < an-1 +8. If ak = 47, how many possibilities are there for the value of k?

(A) Four
(B) Five
(C) Six
(D) Seven
(E) Eight
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Re: PS. Sequence [#permalink] New post 23 Jun 2007, 15:55
kevincan wrote:
In an infinite sequence of integers a1, a2, a3,… , a1 = -30 and for all n > 1, an-1 + 4 <an < an-1 +8. If ak = 47, how many possibilities are there for the value of k?

(A) Four
(B) Five
(C) Six
(D) Seven
(E) Eight



1. The sequence cannot be GP. It has to be AP
2. ak= 47 = -30 + (k-1)d
k-1 = 77/d

Now, d must be such that 4<d<8. Also, 77/d must be an integer. Use these 2 constraints and solve for 77/d.
You can use d = 77/19.......77/10

I can 10 as the answer. Perhaps I m missing something. But I hope you get the general idea.

Someone please rectify my inaccuracy :)
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Re: PS. Sequence [#permalink] New post 23 Jun 2007, 20:00
parsifal wrote:
kevincan wrote:
In an infinite sequence of integers a1, a2, a3,… , a1 = -30 and for all n > 1, an-1 + 4 <an < an-1 +8. If ak = 47, how many possibilities are there for the value of k?

(A) Four
(B) Five
(C) Six
(D) Seven
(E) Eight



1. The sequence cannot be GP. It has to be AP
2. ak= 47 = -30 + (k-1)d
k-1 = 77/d

Now, d must be such that 4<d<8. Also, 77/d must be an integer. Use these 2 constraints and solve for 77/d.
You can use d = 77/19.......77/10

I can 10 as the answer. Perhaps I m missing something. But I hope you get the general idea.

Someone please rectify my inaccuracy :)


I think you are right parsifal. I also get 10.

4<a2 -a1<8
4<a3 -a2<8
.
.
.
4<ak - a(k-1)<8

adding all these equations,

4*(k-1) < ak - a1<8*(k-1)

replacing the values of ak and a1

10.6< k <20.25

k varies from 11 to 20.

Hence 10 values.
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Re: PS. Sequence [#permalink] New post 23 Jun 2007, 23:39
Please ignore

i seem to run into problems when using tags :)

Last edited by parsifal on 23 Jun 2007, 23:43, edited 1 time in total.
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Re: PS. Sequence [#permalink] New post 23 Jun 2007, 23:41
The question says:
In an infinite sequence of integers a1, a2, a3,… , a1 = -30 and for all n > 1, an-1 + 4 <an < an-1 +8 . If ak = 47, how many possibilities are there for the value of k?

In GP, the ratio is constant, the difference will therefore increase beyond 8 or decrease below 4 after the first few terms

Hope that helps.

Whats the OA ?
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 [#permalink] New post 23 Jun 2007, 23:42
Does every sequence of numbers have to be either GP or AP?
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 [#permalink] New post 23 Jun 2007, 23:47
kevincan wrote:
Does every sequence of numbers have to be either GP or AP?


There is a third sequence - Harmonic Progression (HP)
When dealing with problems involving the word "sequence" and an, an-1, AP,GP,HP is the first thing that comes to mind.

But yeah, I never thought of alternatives, ie. the sequence could be simply related by some function and not necessarily be an AP, GP, HP.

I would wait for some more opinions on this thread by other members.
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Re: PS. Sequence [#permalink] New post 24 Jun 2007, 02:22
parsifal wrote:
kevincan wrote:
In an infinite sequence of integers a1, a2, a3,… , a1 = -30 and for all n > 1, an-1 + 4 <an < an-1 +8. If ak = 47, how many possibilities are there for the value of k?

(A) Four
(B) Five
(C) Six
(D) Seven
(E) Eight



1. The sequence cannot be GP. It has to be AP
2. ak= 47 = -30 + (k-1)d
k-1 = 77/d

Now, d must be such that 4<d<8. Also, 77/d must be an integer. Use these 2 constraints and solve for 77/d.
You can use d = 77/19.......77/10

I can 10 as the answer. Perhaps I m missing something. But I hope you get the general idea.

Someone please rectify my inaccuracy :)


Parsifal: Can you pls. explain your solution? I am not understanding it at all.

So we determine that this is an AP not a GP. Then I see that you are using the formula for calc. the nth term of an AP = a+(n-1)d

But how are you using this? How is 47=-30+(k-1)d?
Also why is it that 4<d<8?

I haven't done many AP questions as yet and your explanation will be greatly appreciated. Thanks.
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Re: PS. Sequence [#permalink] New post 24 Jun 2007, 02:24
shoonya wrote:
parsifal wrote:
kevincan wrote:
In an infinite sequence of integers a1, a2, a3,… , a1 = -30 and for all n > 1, an-1 + 4 <an < an-1 +8. If ak = 47, how many possibilities are there for the value of k?

(A) Four
(B) Five
(C) Six
(D) Seven
(E) Eight



1. The sequence cannot be GP. It has to be AP
2. ak= 47 = -30 + (k-1)d
k-1 = 77/d

Now, d must be such that 4<d<8. Also, 77/d must be an integer. Use these 2 constraints and solve for 77/d.
You can use d = 77/19.......77/10

I can 10 as the answer. Perhaps I m missing something. But I hope you get the general idea.

Someone please rectify my inaccuracy :)


I think you are right parsifal. I also get 10.

4<a2 -a1<8
4<a3 -a2<8
.
.
.
4<ak - a(k-1)<8

adding all these equations,

4*(k-1) < ak - a1<8*(k-1)

replacing the values of ak and a1

10.6< k <20.25

k varies from 11 to 20.

Hence 10 values.


Shoonya can you pls explain why 4<a2 -a1<8? Thanks.
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Re: PS. Sequence [#permalink] New post 24 Jun 2007, 04:08
GK_Gmat wrote:
shoonya wrote:
parsifal wrote:
kevincan wrote:
In an infinite sequence of integers a1, a2, a3,… , a1 = -30 and for all n > 1, an-1 + 4 <an < an-1 +8. If ak = 47, how many possibilities are there for the value of k?

(A) Four
(B) Five
(C) Six
(D) Seven
(E) Eight



1. The sequence cannot be GP. It has to be AP
2. ak= 47 = -30 + (k-1)d
k-1 = 77/d

Now, d must be such that 4<d<8. Also, 77/d must be an integer. Use these 2 constraints and solve for 77/d.
You can use d = 77/19.......77/10

I can 10 as the answer. Perhaps I m missing something. But I hope you get the general idea.

Someone please rectify my inaccuracy :)


I think you are right parsifal. I also get 10.

4<a2 -a1<8
4<a3 -a2<8
.
.
.
4<ak - a(k-1)<8

adding all these equations,

4*(k-1) < ak - a1<8*(k-1)

replacing the values of ak and a1

10.6< k <20.25

k varies from 11 to 20.

Hence 10 values.


Shoonya can you pls explain why 4<a2 -a1<8? Thanks.


The stem says that an-1 + 4 <an < an-1 +8.

hence a1 + 4 <a2< a1 + 8

subtract a1 from both sides, and you will get the equation.
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Re: PS. Sequence [#permalink] New post 24 Jun 2007, 06:03
GK_Gmat wrote:
parsifal wrote:
kevincan wrote:
In an infinite sequence of integers a1, a2, a3,… , a1 = -30 and for all n > 1, an-1 + 4 <an < an-1 +8. If ak = 47, how many possibilities are there for the value of k?

(A) Four
(B) Five
(C) Six
(D) Seven
(E) Eight



1. The sequence cannot be GP. It has to be AP
2. ak= 47 = -30 + (k-1)d
k-1 = 77/d

Now, d must be such that 4<d<8. Also, 77/d must be an integer. Use these 2 constraints and solve for 77/d.
You can use d = 77/19.......77/10

I can 10 as the answer. Perhaps I m missing something. But I hope you get the general idea.

Someone please rectify my inaccuracy :)


Parsifal: Can you pls. explain your solution? I am not understanding it at all.

So we determine that this is an AP not a GP. Then I see that you are using the formula for calc. the nth term of an AP = a+(n-1)d

But how are you using this? How is 47=-30+(k-1)d?
Also why is it that 4<d<8?

I haven't done many AP questions as yet and your explanation will be greatly appreciated. Thanks.



Step 1: The given sequence is infinite. So, it can be AP, GP (or HP as mentioned in an earlier post by me). Hence I initially checked if it could be AP, GP or both. I concluded it can only be AP. How ? well, because a GP has a constant ratio and the series is infinite, there will be some n where the difference between 2 successive numbers becomes greater than 8 or less than 4.

Step 2: Now that we know its AP, formulate the AP equation. Note that the question mentions : an-1 + 4 <an < an-1 +8. This means the difference between 2 successive elements is between 4 and 8.. This must be 'd'.

Also, 'k' has to be an integer (you cannot have 10.5th element of sequence)

Thus, we now have enough info to solve it.

Hope that helps!
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Re: PS. Sequence [#permalink] New post 24 Jun 2007, 06:30
parsifal wrote:
GK_Gmat wrote:
parsifal wrote:
kevincan wrote:
In an infinite sequence of integers a1, a2, a3,… , a1 = -30 and for all n > 1, an-1 + 4 <an < an-1 +8. If ak = 47, how many possibilities are there for the value of k?

(A) Four
(B) Five
(C) Six
(D) Seven
(E) Eight



1. The sequence cannot be GP. It has to be AP
2. ak= 47 = -30 + (k-1)d
k-1 = 77/d

Now, d must be such that 4<d<8. Also, 77/d must be an integer. Use these 2 constraints and solve for 77/d.
You can use d = 77/19.......77/10

I can 10 as the answer. Perhaps I m missing something. But I hope you get the general idea.

Someone please rectify my inaccuracy :)


Parsifal: Can you pls. explain your solution? I am not understanding it at all.

So we determine that this is an AP not a GP. Then I see that you are using the formula for calc. the nth term of an AP = a+(n-1)d

But how are you using this? How is 47=-30+(k-1)d?
Also why is it that 4<d<8?

I haven't done many AP questions as yet and your explanation will be greatly appreciated. Thanks.



Step 1: The given sequence is infinite. So, it can be AP, GP (or HP as mentioned in an earlier post by me). Hence I initially checked if it could be AP, GP or both. I concluded it can only be AP. How ? well, because a GP has a constant ratio and the series is infinite, there will be some n where the difference between 2 successive numbers becomes greater than 8 or less than 4.

Step 2: Now that we know its AP, formulate the AP equation. Note that the question mentions : an-1 + 4 <an < an-1 +8. This means the difference between 2 successive elements is between 4 and 8.. This must be 'd'.

Also, 'k' has to be an integer (you cannot have 10.5th element of sequence)

Thus, we now have enough info to solve it.

Hope that helps!


Parsifal, I would like to disagree with you on this. This is necessarily not an AP. In an AP, the difference between the two consecutive terms is constant, here it's not the case. The difference varies from 5 to 7 and hence there can be different numbers of terms that satisfy the condition.
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 [#permalink] New post 24 Jun 2007, 07:27
Shoonya:
Well, we are talking of 10 different APs here. The question indirectly asks you to find the number of different sequences.
Note that an AP is completely defined by a1 and d. a1 is given. Thus, we need to find the number of possible d's.

hope that clarifies!
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 [#permalink] New post 24 Jun 2007, 08:33
This need not be an AP- a sequence is simply a list of numbers in a definite order
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 [#permalink] New post 24 Jun 2007, 09:34
kevincan wrote:
This need not be an AP- a sequence is simply a list of numbers in a definite order


I agree with Kevin. This is how I look at this series.
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 [#permalink] New post 24 Jun 2007, 13:16
kevincan wrote:
This need not be an AP- a sequence is simply a list of numbers in a definite order


I agree with this, and the answer is 5. Trying to use too many mathematical principles can kill a gmat question. Sometimes, it's just brute force that gets the right answer. I prefer to do a sequence out a few times, look for patterns, and then go to the answer.

Look at this question. It says that a1 is -30, so using the given formula, a2 must be either -25, -24, -23. All good and well, but now we have three numbers to go off of for a2. If we try it for all of them, we see that a2 can be either -20, -19, -18, -17, -16. There are now 5 possibilities.

Keep doing it out for a few more. You will see that each term grows in the number of possibilities by 2. So a3 will have 7 possible answers, a4 will have 9, etc.

Then you just need to figure out the first possible number for each term, which is always based on the previous first possible number, and then get a range.

So, since a3 started with -20, the smallest number for a4 is -15, and there are 7 terms, so the range is -15 to -9.

a5: smalles term, based on a3, is -10. 9 terms, so range is -10 to -2.

a6: 11 terms, range: -5 to +5

a7: 13 terms, range: 0 to 12

a8: 15 terms, range 5 to 19

a9: 17 terms, range 10 to 26

a10: 19 terms, range 15 to 33

a11: 21 terms, range 20 to 40

a12: 23 terms, range 25 to 47 <-- this is the first k!

a13: 25 terms, range 30 to 54 <-- 2nd k

a14: 27 terms, starting with 35 <-- 3rd k

a15: 29 terms, starting with 40 <-- 4th k

a16: 31 terms, starting with 45 <-- 5th k

a17: 33 terms, starting with 50 <-- no more k!

So there are 6 possible k's that work, thus the answer is 5.

That said, this is too much work for the gmat. I would never expect to see it there.
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 [#permalink] New post 24 Jun 2007, 14:20
Ian, this is far too much work! Remember that GMAT questions do not require brute force
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 [#permalink] New post 24 Jun 2007, 14:27
kevincan wrote:
Ian, this is far too much work! Remember that GMAT questions do not require brute force


Agreed! That's why I don't think this is a real GMAT question.

But it's still good to get practice on thinking these things out. If the question said that Ak= 11, then we'd only have gotten up to the 9th terms, as oposed to the 17th, and then it's more likely to be on the test. Also, it didn't have to start at -30. That also adds time.

I think, in general, that a gmat question could legitimately require someone to use brute force on sequences up to about the 7th or 8th term. After that you need to find a pattern, but before that you usually have to just start with a1 and use the formula to get to a7 or a8.

So far, the only other way people have come up with, a highly mathematical way, which assumed the sequence was either arithmetic or geometric, was wrong. The answer is 5, not 10.

Regarding brute force, there are just some times that there is no other option, and there are also times when brute force is all someone has who cannot think of the most sophisticated method. It is better to spend a minute and a half doing something the long way than spending 2 minutes trying to think of the sophisticated way, when it may not come to you.

Just my opinions...
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 [#permalink] New post 24 Jun 2007, 14:43
We know that each term is either 5,6 or 7 greater than the previous term

Thus 5(k-1) is less than or equal to ak - (-30) which is less than or equal to 7(k-1)
We know that ak = 47
Thus 7(k-1) >= 77 i.e. k is greater than or equal to 12
5(k - 1) <= 77 i.e k is less than or equal to 16

Thus k can be either 12, 13, 14, 15, 16

OA = B

Thanks for all who tried. I hope we see that not all sequences are AP, GP ...
Ian, I will change the numbers in my question to make it a bit more amenable to brute force. Thanks for the comment
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 [#permalink] New post 24 Jun 2007, 15:01
kevincan wrote:
We know that each term is either 5,6 or 7 greater than the previous term

Thus 5(k-1) is less than or equal to ak - (-30) which is less than or equal to 7(k-1)
We know that ak = 47
Thus 7(k-1) >= 77 i.e. k is greater than or equal to 12
5(k - 1) <= 77 i.e k is less than or equal to 16

Thus k can be either 12, 13, 14, 15, 16

OA = B

Thanks for all who tried. I hope we see that not all sequences are AP, GP ...
Ian, I will change the numbers in my question to make it a bit more amenable to brute force. Thanks for the comment


I like your answer here. Did you write the question? It's a good one.
  [#permalink] 24 Jun 2007, 15:01
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