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In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3.

(2) The second-to-last term is 3^10.

I think the ans should be D. But OA is different.

In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3 --> sequence is: 3, 9, 27, 81, ... so the fourth term is 81. Sufficient.

(2) The second-to-last term is 3^10 --> since we don't know how many terms are there in the sequence then we don't know which term is second-to-last. For example: if it's third term then fourth (and last) will be 3^11, if it's fifth term then the fourth term is 3^9, ... Not sufficient.

In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3.

(2) The second-to-last term is 3^10.

I think the ans should be D. But OA is different.

regarding the second statement. Someone could interpret the second-to-last term as the ratio between the second and the last term.

Since we know that it is an exponential expression dividing the second with the last term should result a fraction smaller than 1. Therefore, someone could assume that the second statement is wrong.

is my reasoning valid?

Responding to a pm.

If it were the case it would have been something like "the ratio of second to last term is ..."

Also on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other. So if on the GMAT your interpretation of the statements leads you to conclude that the statements are impossible/incorrect or contradict each other then the case would be that your interpretation is wrong not the statements. _________________

In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3.

(2) The second-to-last term is 3^10.

I think the ans should be D. But OA is different.

In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3 --> sequence is: 3, 9, 27, 81, ... so the fourth term is 81. Sufficient.

(2) The second-to-last term is 3^10 --> since we don't know how many terms are there in the sequence then we don't know which term is second-to-last. For example: if it's third term then fourth (and last) will be 3^11, if it's fifth term then the fourth term is 3^9, ... Not sufficient.

Answer: A.

I had to think about that for a bit before I could sincerely agree with you, Bunuel. Thanks! _________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done." -Nelson Mandela

In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3.

(2) The second-to-last term is 3^10.

I think the ans should be D. But OA is different.

In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3 --> sequence is: 3, 9, 27, 81, ... so the fourth term is 81. Sufficient.

(2) The second-to-last term is 3^10 --> since we don't know how many terms are there in the sequence then we don't know which term is second-to-last. For example: if it's third term then fourth (and last) will be 3^11, if it's fifth term then the fourth term is 3^9, ... Not sufficient.

Answer: A.

You are truly awesome .. with DS.... i kinda assumed it to be 2nd term when said 2nd to last assuming that there r only 4 terms...thanks bunuel...

Re: In a sequence of terms in which each term is three times the [#permalink]

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13 Jan 2012, 16:39

Bunuel wrote:

mariyea wrote:

In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3.

(2) The second-to-last term is 3^10.

I think the ans should be D. But OA is different.

regarding the second statement. Someone could interpret the second-to-last term as the ratio between the second and the last term.

Since we know that it is an exponential expression dividing the second with the last term should result a fraction smaller than 1. Therefore, someone could assume that the second statement is wrong.

Re: In a sequence of terms in which each term is three times the [#permalink]

Show Tags

24 Oct 2014, 08:39

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: In a sequence of terms in which each term is three times the [#permalink]

Show Tags

14 Jan 2016, 12:06

Bunuel wrote:

mariyea wrote:

In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3.

(2) The second-to-last term is 3^10.

I think the ans should be D. But OA is different.

In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3 --> sequence is: 3, 9, 27, 81, ... so the fourth term is 81. Sufficient.

(2) The second-to-last term is 3^10 --> since we don't know how many terms are there in the sequence then we don't know which term is second-to-last. For example: if it's third term then fourth (and last) will be 3^11, if it's fifth term then the fourth term is 3^9, ... Not sufficient.

Answer: A.

Hi Bunuel, since the question states that each term is 3 times another term, it means that for the nth term, the value will be 3^n. So I assumed that the 10th term is 3^10, last term is 3^11 and 4th term is 3^4. What was the flaw in my logic? Thanks in advance.

Hi Bunuel, since the question states that each term is 3 times another term, it means that for the nth term, the value will be 3^n. So I assumed that the 10th term is 3^10, last term is 3^11 and 4th term is 3^4. What was the flaw in my logic? Thanks in advance.

You are not interpreting the text in red above correctly.

When you say "3 times the other term", it means that if the 1st term is a, then the 2nd term is 3a, 3rd term is 3a*3=9a etc. Thus the sequence becomes

a,3a,9a,27a .... and NOT 3^n as you have mentioned.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3.

(2) The second-to-last term is 3^10.

Modify the original condition and the question and suppose the sequence A_n. Then it becomes A_(n+1)=3A_(n) and once you figure out A_(1), you can figure out everything. So there is 1 variable, which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. For 1), A_(1)=3 -> A_(2)=3^2, A_(3)=3^3, A_(4)=3^4, which is unique and sufficient. For 2), you can’t figure out where the last term comes, which is not sufficient. Therefore, the answer is A.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E. _________________

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