Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Sequence OG, DS [#permalink]
16 Jun 2009, 04:49

1

This post received KUDOS

Answer is C. I dont understand how it could be B if K must not be 0? The set has 14 numbers. "A8" is the 7th of those 14 numbers. In any with an even amount of numbers there is no middle number and you are therefore not able to decide if there are 6 or 7 numbers >10.

It does not matter if they give you A7 and 1<= n <=14, A8 and 2<= n <=15 or A86 and 80<= n <=93. As long as the set consists of an even amount of numbers there is no middle number.

Re: Sequence OG, DS [#permalink]
22 Aug 2009, 21:30

An = An-1 + K for K! = 0 would be arithmetic expression with K as the difference. Hence it can be concluded from second expression(A8=10) that in a decreasing sequence number , 7th term is the closest term to the value of 10. 7 terms have value more than 10.

more detailed analysis (though Statement 2 is enough for conclusion) putting in A1 = 10 and A8 = 24,

A8 = A1 + (8-1)K -- ( n term of the sequence = A1 + (n-1)K ) 10 = A1 + 7K 10 = 24 + 7K K = - 2

The value of 7th term is A7 = A1 + 6k = 24 - 12 = 12 > 10 . ----------------------

Statement 1 is not sufficient enough / Statement 2 is sufficient Statement 1 + 2 = confirm the values further

Re: Sequence OG, DS [#permalink]
23 Aug 2009, 03:41

rohansherry wrote:

IMO C: , pls tell us which ques from OG is it. Guys the OA mentioned above is B. PLs someone explain it.

Answer should B because:

a8= 10. Hence, a9=10+k, a10 = 10 +2k and so on....Also, a7= 10-k, a6=10-2k and so on... Now, if K is positive then 7 terms on the right of a8 are >10 and if K is negative then 7 terms on the left side are >10.

Re: Sequence OG, DS [#permalink]
23 Aug 2009, 08:04

Assumption: K is a constant according the defined sequence nth term. Since A1 = 24 and A8 = 10, K is negative in a decreasing sequence. 7 terms on the left side are >10

If K is not a constant but some function of the terms like K = n^((-1)^n) which will give fractions in case of odd terms and whole integers in case of even. or K = (-1)^n in which case the sequence can have positive and negative numbers based on the value of n. This will lead the solution being E as both statements are insufficient. But I think that assumption made about K being constant is correct according to the definition of a sequence.

Economist wrote:

rohansherry wrote:

IMO C: , pls tell us which ques from OG is it. Guys the OA mentioned above is B. PLs someone explain it.

Answer should B because:

a8= 10. Hence, a9=10+k, a10 = 10 +2k and so on....Also, a7= 10-k, a6=10-2k and so on... Now, if K is positive then 7 terms on the right of a8 are >10 and if K is negative then 7 terms on the left side are >10.

Re: Sequence OG, DS [#permalink]
23 Aug 2009, 08:46

Economist wrote:

rohansherry wrote:

IMO C: , pls tell us which ques from OG is it. Guys the OA mentioned above is B. PLs someone explain it.

Answer should B because:

a8= 10. Hence, a9=10+k, a10 = 10 +2k and so on....Also, a7= 10-k, a6=10-2k and so on... Now, if K is positive then 7 terms on the right of a8 are >10 and if K is negative then 7 terms on the left side are >10.

Why don't you consider k=0? If k=0, then a1=a2=...=a8=a9=...=a15=10...so no inetger is greater than 10...so the answer could be: 1) zero if k= 0 2) 7 if K is either positive or negative

Combining statements 1 and 2, it can be concluded that k is negative. Sufficient. There are 7 numbers greater than 10.

Can someone please explain how there are 15 terms in this set? I can only see 14 if n is truly >= 2 and <=15.

I edited the question above, there was one part missing.

We have a sequence of fifteen terms (actually this sequence is arithmetic progression). As k is nonzero, all elements would be different and the median would be the eighth term, a_8. This means that 7 terms will be less than a_8 and 7 terms will be more than a_8. Note here that it doesn't matter whether k is positive or negative:

if k is positive, we'll get an ascending sequence and the terms from from a_1 to a_7 will be less than a_8 and terms from a_9 to a_{15} will be more than a_8;

if k is negative, we'll get an descending sequence and the terms from from a_1 to a_7 will be more than a_8 and terms from a_9 to a_{15} will be less than a_8.

Statement (1) is giving the value of a_1, but since we don't know k, we can not say how many terms are more than 10: it can vary from 1 (only a_1=24>10, if k<=-14) to 15 (if k is positive for instance).

Statement (2) is saying that a_8=10. As we discussed above, a_8 is median value and for any value of k, 7 terms will be more than a_8=10 and 7 terms will be less than a_8=10. Hence this statement is sufficient.