In an infinite sequence of integers a1, a2, a3, , a1 = -30 : PS Archive - Page 2
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# In an infinite sequence of integers a1, a2, a3, , a1 = -30

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24 Jun 2007, 15:05
ian7777 wrote:
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.

Yeah, feel free to use it- there aren't enough good sequence questions and I teach the GMAT online. I just wanted to try it out before I incorporate it
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24 Jun 2007, 15:51
Kevin, I hear you. I'm trying to figure out why using 4<a(n) - a(n-1)<8 gives us a wrong value and 5<=a(n) -a(n-1),=7 gives us the right one. We both approached it the same way. It's just that you divided 77 by 5 and 8 to get the number and I and parsifal divided it by 4 and 8. We also should have gotten the right answer using our equation. What is the fundamental error here?
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24 Jun 2007, 16:12
Hey Kevin,

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

The condition highlighted forces this sequence to be an AP. In that case answer should be "one" since n should be an integer value and is 12
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24 Jun 2007, 19:52
kevincan wrote:
We know that each term is either 5,6 or 7 greater than the previous term

Why not 4.5, 5.5, 6.5, 7.7 ..... The question doesnt say that it has to be greater by an integral value.

kevincan wrote:
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

If you use difference as 7.7, then k can be 11. Similarly other values of k possible. As solved in my earlier post, there can be a total of 10 values.

kevincan wrote:
Thanks for all who tried. I hope we see that not all sequences are AP, GP

But this sequence is. In fact, indirectly you yourself have used the formulas that are typical of APs.
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25 Jun 2007, 00:02
vijay2001 wrote:
Hey Kevin,

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

The condition highlighted forces this sequence to be an AP. In that case answer should be "one" since n should be an integer value and is 12

It need not be an AP. As Ian remarked, a2 could be -25 and a3 = - 19
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25 Jun 2007, 08:00
kevincan wrote:
Remember that the terms must be integers

mea culpa!
peccavi !

I completely overlooked that. Sorry for all the ensuing confusion.
25 Jun 2007, 08:00

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