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The sequence a(n) is defined so that, for all n is greater [#permalink]

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08 Mar 2012, 22:14

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The sequence \(a_n\) is defined so that, for all \(n\) is greater than or equal to 3, \(a_n\) is the greater of \(a_{n-2} +1\) and \(a_{n-1}\). (If the two quantities are the same, then \(a_n\) is equal to either of them.) Which of the following values of \(a_1\) and \(a_2\) will produce a sequence in which no value is repeated?

A. \(a_1=-1\), \(a_2=-1.5\) B. \(a_1=-1\), \(a_2=1\) C. \(a_1=1\), \(a_2=-1\) D. \(a_1=1\), \(a_2=1.5\) E. \(a_1=1.5\), \(a_2=1\)

I am having trouble understanding what the question is asking and also solving it

The sequence a(n) is defined so that, for all n is greater than or equal to 3, a(n) is the greater of (a(n-2) +1) and (an-1). (If the two quantities are the same, then an is equal to either of them.) Which of the following values of a1 and a2 will produce a sequence in which no value is repeated?

a) a1=-1, a2=-1.5 b) a1=-1, a2=1 c) a1=1, a2=-1 d) a1=1, a2=1.5 e) a1=1.5, a2=1

I am having trouble understanding what the question is asking and also solving it

Take one line of the question at a time and try to make sense of it. Sequence questions seem daunting due to all the subscripts but they are pretty straight forward, generally.

Given: \(a_n\) = Greater of \((a_{n-2} + 1, a_{n-1})\) (n is 3 or greater) This means that starting from the third term, every term is the greater of (one more than previous to previous term, the previous term)

If we want that every term in the sequence should be unique, \(a_n\) should not be equal to the previous term. It should be equal to 'one more than previous to previous term'

So what can you deduce about \(a_1\) and \(a_2\)? 1. \(a_1 + 1\) should be greater than \(a_2\) so that \(a_3 \neq a_2\). Reject option (B)

2. To ensure that \(a_4 \neq a_3\), \(a_4 = a_2 + 1\). Therefore, \(a_2 + 1 > a_3\)

Option (A) \(a_3\) = 0 which is greater than \(a_2 + 1 (= -0.5)\) so reject it. Option (C) \(a_3\) = 2 which is greater than \(a_2 + 1 (= 0)\) so reject it. Option (E) \(a_3\) = 2.5 which is greater than \(a_2 + 1 (= 2)\) so reject it.

Answer must be option (D). \(a_3\) = 2 which is less than \(a_2 + 1\).

OR if you want to think the logical way, realize that the first term must be smaller than the second term but the difference between them should be less than 1 (so that when 1 is added, it becomes more than the second term). If you understand this, then you can quickly jump to option (D) _________________

The sequence a(n) is defined so that, for all n is greater than or equal to 3, a(n) is the greater of (a(n-2) +1) and (an-1). (If the two quantities are the same, then an is equal to either of them.) Which of the following values of a1 and a2 will produce a sequence in which no value is repeated?

a) a1=-1, a2=-1.5 b) a1=-1, a2=1 c) a1=1, a2=-1 d) a1=1, a2=1.5 e) a1=1.5, a2=1

I am having trouble understanding what the question is asking and also solving it

Take one line of the question at a time and try to make sense of it. Sequence questions seem daunting due to all the subscripts but they are pretty straight forward, generally.

Given: \(a_n\) = Greater of \((a_{n-2} + 1, a_{n-1})\) (n is 3 or greater) This means that starting from the third term, every term is the greater of (one more than previous to previous term, the previous term)

If we want that every term in the sequence should be unique, \(a_n\) should not be equal to the previous term. It should be equal to 'one more than previous to previous term'

So what can you deduce about \(a_1\) and \(a_2\)? 1. \(a_1 + 1\) should be greater than \(a_2\) so that \(a_3 \neq a_2\). Reject option (B)

2. To ensure that \(a_4 \neq a_3\), \(a_4 = a_2 + 1\). Therefore, \(a_2 + 1 > a_3\)

Option (A) \(a_3\) = 0 which is greater than \(a_2 + 1 (= -0.5)\) so reject it. Option (C) \(a_3\) = 2 which is greater than \(a_2 + 1 (= 0)\) so reject it. Option (E) \(a_3\) = 2.5 which is greater than \(a_2 + 1 (= 2)\) so reject it.

Answer must be option (D). \(a_3\) = 2 which is less than \(a_2 + 1\).

OR if you want to think the logical way, realize that the first term must be smaller than the second term but the difference between them should be less than 1 (so that when 1 is added, it becomes more than the second term). If you understand this, then you can quickly jump to option (D)

@karishma To be honest what is probability of such questions landing up on your gmat and on what level ... I think, I would just flip the computer table and walk out rather than solve such types

@karishma To be honest what is probability of such questions landing up on your gmat and on what level ... I think, I would just flip the computer table and walk out rather than solve such types

The question is not tough Utkarsh. I would suggest you to go one step at a time and try to figure it out. Such questions can be a part of actual GMAT and it's just a 700+, not exceptionally over the top. It looks complicated but most of the steps would be kind of intuitive after some practice. GMAT excels at testing simple concepts wrapped in a twisted package. _________________

The sequence a(n) is defined so that, for all n is greater than or equal to 3, a(n) is the greater of (a(n-2) +1) and (an-1). (If the two quantities are the same, then an is equal to either of them.) Which of the following values of a1 and a2 will produce a sequence in which no value is repeated?

a) a1=-1, a2=-1.5 b) a1=-1, a2=1 c) a1=1, a2=-1 d) a1=1, a2=1.5 e) a1=1.5, a2=1

I am having trouble understanding what the question is asking and also solving it

Take one line of the question at a time and try to make sense of it. Sequence questions seem daunting due to all the subscripts but they are pretty straight forward, generally.

Given: \(a_n\) = Greater of \((a_{n-2} + 1, a_{n-1})\) (n is 3 or greater) This means that starting from the third term, every term is the greater of (one more than previous to previous term, the previous term)

If we want that every term in the sequence should be unique, \(a_n\) should not be equal to the previous term. It should be equal to 'one more than previous to previous term'

So what can you deduce about \(a_1\) and \(a_2\)? 1. \(a_1 + 1\) should be greater than \(a_2\) so that \(a_3 \neq a_2\). Reject option (B)

2. To ensure that \(a_4 \neq a_3\), \(a_4 = a_2 + 1\). Therefore, \(a_2 + 1 > a_3\)

Option (A) \(a_3\) = 0 which is greater than \(a_2 + 1 (= -0.5)\) so reject it. Option (C) \(a_3\) = 2 which is greater than \(a_2 + 1 (= 0)\) so reject it. Option (E) \(a_3\) = 2.5 which is greater than \(a_2 + 1 (= 2)\) so reject it.

Answer must be option (D). \(a_3\) = 2 which is less than \(a_2 + 1\).

OR if you want to think the logical way, realize that the first term must be smaller than the second term but the difference between them should be less than 1 (so that when 1 is added, it becomes more than the second term). If you understand this, then you can quickly jump to option (D)

Hi Karishma,

Thanks for your blog...I have been trying to solve every problem without using a pen per your advice and I am getting better at the quant section as a result.

Here is where I am confused about the above problem:

"If we want that every term in the sequence should be unique, \(n\) should not be equal to the previous term. It should be equal to 'one more than previous to previous term'"

I'm reading the rule as \(A(n)\) is greater than \(A(n-2) + 1\). Above, you say they must be equal (at least that is how I'm reading it). Overall, the wording of this problem is awkward to me. I've never seen the phrase "...is the greater of". That just means it is great than, right?

"If we want that every term in the sequence should be unique, \(n\) should not be equal to the previous term. It should be equal to 'one more than previous to previous term'"

I'm reading the rule as \(A(n)\) is greater than \(A(n-2) + 1\). Above, you say they must be equal (at least that is how I'm reading it). Overall, the wording of this problem is awkward to me. I've never seen the phrase "...is the greater of". That just means it is great than, right?

There are 2 diff things:

1. 'Is greater than' x is greater than 4 and 5. This means x is a number greater than 5.

2. 'Is greater of' x is greater of 4 and 5. This means x = 5. Look at it this way: x = Greater of (4, 5) First you find that which number is greater out of 4 and 5. x will be equal to that number.

x = Greater of (Last to last term + 1, Last term) means find which is greater 'Last to last term + 1' or 'Last term'. X will be equal to that. _________________

Re: The sequence a(n) is defined so that, for all n is greater [#permalink]

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21 Feb 2014, 12:35

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: The sequence a(n) is defined so that, for all n is greater [#permalink]

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03 Jul 2014, 09:41

VeritasPrepKarishma wrote:

shawndx wrote:

The sequence a(n) is defined so that, for all n is greater than or equal to 3, a(n) is the greater of (a(n-2) +1) and (an-1). (If the two quantities are the same, then an is equal to either of them.) Which of the following values of a1 and a2 will produce a sequence in which no value is repeated?

a) a1=-1, a2=-1.5 b) a1=-1, a2=1 c) a1=1, a2=-1 d) a1=1, a2=1.5 e) a1=1.5, a2=1

I am having trouble understanding what the question is asking and also solving it

Take one line of the question at a time and try to make sense of it. Sequence questions seem daunting due to all the subscripts but they are pretty straight forward, generally.

Given: \(a_n\) = Greater of \((a_{n-2} + 1, a_{n-1})\) (n is 3 or greater) This means that starting from the third term, every term is the greater of (one more than previous to previous term, the previous term)

If we want that every term in the sequence should be unique, \(a_n\) should not be equal to the previous term. It should be equal to 'one more than previous to previous term'

So what can you deduce about \(a_1\) and \(a_2\)? 1. \(a_1 + 1\) should be greater than \(a_2\) so that \(a_3 \neq a_2\). Reject option (B)

2. To ensure that \(a_4 \neq a_3\), \(a_4 = a_2 + 1\). Therefore, \(a_2 + 1 > a_3\)

Option (A) \(a_3\) = 0 which is greater than \(a_2 + 1 (= -0.5)\) so reject it. Option (C) \(a_3\) = 2 which is greater than \(a_2 + 1 (= 0)\) so reject it. Option (E) \(a_3\) = 2.5 which is greater than \(a_2 + 1 (= 2)\) so reject it.

Answer must be option (D). \(a_3\) = 2 which is less than \(a_2 + 1\).

OR if you want to think the logical way, realize that the first term must be smaller than the second term but the difference between them should be less than 1 (so that when 1 is added, it becomes more than the second term). If you understand this, then you can quickly jump to option (D)

KARISHMA (SORRY FOR CAPS JUST TRYING TO HIGHLIGHT WHERE MY QUESTION IS):

WHEN YOU SAY N IS 3 OR GREATER TOWARDS THE BEGINNING OF YOUR ANSWER HOW EXACTLY DID YOU DEDUCE THAT? SECONDLY WHEN YOU ARE TESTING ANSWER CHOICES HOW EXACTLY DO YOU GET THE VALUE OF A3? FOR EXAMPLE WHEN YOU TEST CHOICE C HOW DID YOU GET A3 = 2?

Re: The sequence a(n) is defined so that, for all n is greater [#permalink]

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03 Jul 2014, 19:38

Expert's post

sagnik2422 wrote:

KARISHMA (SORRY FOR CAPS JUST TRYING TO HIGHLIGHT WHERE MY QUESTION IS):

WHEN YOU SAY N IS 3 OR GREATER TOWARDS THE BEGINNING OF YOUR ANSWER HOW EXACTLY DID YOU DEDUCE THAT? SECONDLY WHEN YOU ARE TESTING ANSWER CHOICES HOW EXACTLY DO YOU GET THE VALUE OF A3? FOR EXAMPLE WHEN YOU TEST CHOICE C HOW DID YOU GET A3 = 2?

THANKS

Since all previous quotes have a lighter background, it is easy to see where your question is so don't worry about that.

WHEN YOU SAY N IS 3 OR GREATER TOWARDS THE BEGINNING OF YOUR ANSWER HOW EXACTLY DID YOU DEDUCE THAT?

It is given in the question: "The sequence a(n) is defined so that, for all n is greater than or equal to 3".

SECONDLY WHEN YOU ARE TESTING ANSWER CHOICES HOW EXACTLY DO YOU GET THE VALUE OF A3? FOR EXAMPLE WHEN YOU TEST CHOICE C HOW DID YOU GET A3 = 2?

We know that An is greater of A(n-2) + 1 and A(n-1). This means An will be either (A(n-2)+ 1) or A(n-1), whichever is greater! Now use options:

a) a1=-1, a2=-1.5

A3 = Greater of (A(1)+1, A(2)) A3 = Greater of (-1+1, -1.5) A3 = Greater of (0, -1.5) A3 = 0

c) a1=1, a2=-1

A3 = Greater of (A(1)+1, A(2)) A3 = Greater of (1+1, -1) A3 = Greater of (2, -1) A3 = 2

Re: The sequence a(n) is defined so that, for all n is greater [#permalink]

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16 Mar 2015, 02:02

shawndx wrote:

The sequence \(a_n\) is defined so that, for all \(n\) is greater than or equal to 3, \(a_n\) is the greater of \(a_{n-2} +1\) and \(a_{n-1}\). (If the two quantities are the same, then \(a_n\) is equal to either of them.) Which of the following values of \(a_1\) and \(a_2\) will produce a sequence in which no value is repeated?

A. \(a_1=-1\), \(a_2=-1.5\) B. \(a_1=-1\), \(a_2=1\) C. \(a_1=1\), \(a_2=-1\) D. \(a_1=1\), \(a_2=1.5\) E. \(a_1=1.5\), \(a_2=1\)

I am having trouble understanding what the question is asking and also solving it

I solved the first option to get a hang of what kind of series comes out... Quickly figured out that only in D it is possible \(a_{n-2} +1\) is always greater than \(a_{n-1}\) bcoz both entities are positive and \(a_2=1.5\) > \(a_1=1\) _________________

Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time.

I hated every minute of training, but I said, 'Don't quit. Suffer now and live the rest of your life as a champion.-Mohammad Ali

Re: The sequence a(n) is defined so that, for all n is greater [#permalink]

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21 Mar 2015, 08:22

VeritasPrepKarishma wrote:

shawndx wrote:

The sequence a(n) is defined so that, for all n is greater than or equal to 3, a(n) is the greater of (a(n-2) +1) and (an-1). (If the two quantities are the same, then an is equal to either of them.) Which of the following values of a1 and a2 will produce a sequence in which no value is repeated?

a) a1=-1, a2=-1.5 b) a1=-1, a2=1 c) a1=1, a2=-1 d) a1=1, a2=1.5 e) a1=1.5, a2=1

I am having trouble understanding what the question is asking and also solving it

Take one line of the question at a time and try to make sense of it. Sequence questions seem daunting due to all the subscripts but they are pretty straight forward, generally.

Given: \(a_n\) = Greater of \((a_{n-2} + 1, a_{n-1})\) (n is 3 or greater) This means that starting from the third term, every term is the greater of (one more than previous to previous term, the previous term)

If we want that every term in the sequence should be unique, \(a_n\) should not be equal to the previous term. It should be equal to 'one more than previous to previous term'

So what can you deduce about \(a_1\) and \(a_2\)? 1. \(a_1 + 1\) should be greater than \(a_2\) so that \(a_3 \neq a_2\). Reject option (B)

2. To ensure that \(a_4 \neq a_3\), \(a_4 = a_2 + 1\). Therefore, \(a_2 + 1 > a_3\)

Option (A) \(a_3\) = 0 which is greater than \(a_2 + 1 (= -0.5)\) so reject it. Option (C) \(a_3\) = 2 which is greater than \(a_2 + 1 (= 0)\) so reject it. Option (E) \(a_3\) = 2.5 which is greater than \(a_2 + 1 (= 2)\) so reject it.

Answer must be option (D). \(a_3\) = 2 which is less than \(a_2 + 1\).

OR if you want to think the logical way, realize that the first term must be smaller than the second term but the difference between them should be less than 1 (so that when 1 is added, it becomes more than the second term). If you understand this, then you can quickly jump to option (D)

I have noticed that manahattan gmat questions are very time intensive. i would rather build the series for each of the given options A-E , atleast 2 elements . _________________

Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

Re: The sequence a(n) is defined so that, for all n is greater [#permalink]

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18 May 2016, 19:52

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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