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In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k

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In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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Updated on: 22 May 2013, 03:34
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Given a sequence: $$a_1, \ a_2, \ a_3, \ ... \ a_{14}, \ a_{15}$$

In the sequence shown, $$a_n = a_{n-1}+k$$, where $$2\leq{n}\leq{15}$$ and $$k$$ is a nonzero constant. How many of the terms in the sequence are greater than 10?

(1) $$a_1= 24$$
(2) $$a_8= 10$$

Originally posted by mushyyy on 15 Jan 2012, 10:28.
Last edited by Bunuel on 22 May 2013, 03:34, edited 3 times in total.
Edited the question
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In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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15 Jan 2012, 10:37
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34
mushyyy wrote:
In the sequence shown, a(n) = a(n-1)+k, where 2<=n<=15 and k is a nonzero constant. How many of the
terms in the sequence are greater than 10?
(1) a1 = 24
(2) a8 = 10

OA:B

I think answer is actually C, but before explain my view, I'd like to know your own opinion.

Given a sequence: $$a_1, \ a_2, \ a_3, \ ... \ a_{14}, \ a_{15}$$

In the sequence shown, $$a_n = a_{n-1}+k$$, where $$2\leq{n}\leq{15}$$ and $$k$$ is a nonzero constant. How many of the terms in the sequence are greater than 10?

(1) $$a_1= 24$$
(2) $$a_8= 10$$

We have a sequence of fifteen terms (actually this sequence is an 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 a 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 the value of $$k$$, 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.

Hope it helps.

P.S. Please do not reword the questions when posting (you omitted the crucial part: the sequence itself).
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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05 Jun 2013, 04:17
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2
In the sequence shown, a(n) = a(n-1)+k, where 2<=n<=15 and k is a nonzero constant. How many of the
terms in the sequence are greater than 10?
(1) a1 = 24
(2) a8 = 10

Quite tricky question. This kind of questions are very good excersise to keep you allert that unless stated number could be anything including positive negative.

1 st) a1=24 means that next number a2=24+some number let say 100=124, then a3=124+100=224 etc. So all the numbers in the sequence are greater than 10. But stop! If K is negative number, then a2=24 - some number let say -100=-76, then a3=-76-100=-176. In this case only a1 is greater than 10. Statement is not sufficient to make final answer.

2 st.) a8=10 means that half of the numbers will be greater than a8 and half will be less than. Although we do not know which half, becasue we don't know the sign of the K, but we definately know that 7 numbers will be greater than 10, that was the question all about. So statemnt 2 is sufficient. B
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15 Jan 2012, 10:50
Bunuel wrote:
mushyyy wrote:
In the sequence shown, a(n) = a(n-1)+k, where 2<=n<=15 and k is a nonzero constant. How many of the
terms in the sequence are greater than 10?
(1) a1 = 24
(2) a8 = 10

OA:B

I think answer is actually C, but before explain my view, I'd like to know your own opinion.

Given a sequence: $$a_1, \ a_2, \ a_3, \ ... \ a_{14}, \ a_{15}$$

In the sequence shown, $$a_n = a_{n-1}+k$$, where $$2\leq{n}\leq{15}$$ and $$k$$ is a nonzero constant. How many of the terms in the sequence are greater than 10?

(1) $$a_1= 24$$
(2) $$a_8= 10$$

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 the value of $$k$$, 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.

Hope it helps.

P.S. Please do not reword the questions when posting (you omitted the crucial part: the sequence itself).

yes this was the OA.. but as you see in the problem, n must be equal or greater than 2, so A(1) is impossible...am I wrong?
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15 Jan 2012, 11:00
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mushyyy wrote:
yes this was the OA.. but as you see in the problem, n must be equal or greater than 2, so A(1) is impossible...am I wrong?

Yes. First of all the stem shows you a sequence and there is a first term present (naturally), moreover (1) directly tells us the value of the first term.

$$a_n = a_{n-1}+k$$, where $$2\leq{n}\leq{15}$$ means that we are given the formula to calculate nth term of the sequence starting from the second term: --> $$a_2 = a_{1}+k$$ (it's a common way to give the formula of a sequence).

Hope it's clear.
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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15 Jan 2012, 11:04
okay..I thought the sequence couldn't be A1-A15 and so the middle term wasn't A8..
thanks..+kudos
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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04 Jun 2013, 05:06
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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23 Jun 2013, 17:30
1
I completely missed this question because when I read the 2 =< N =< 15 section, I thought that that was the terms of the sequence, which would give fourteen entries and yield choice C.

I see now that the 2 =< N =< 15 is the conditional for where to apply the formula, not the length of the sequence.

Clever clever gmat...
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04 Sep 2013, 03:02
1
I am yet to meet someone who has such ability to write lucid answers for seemingly tough questions.

Bunuel, you deserve something big, very big in field of Mathematics.
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11 Sep 2013, 05:50
Bunuel wrote:
mushyyy wrote:
In the sequence shown, a(n) = a(n-1)+k, where 2<=n<=15 and k is a nonzero constant. How many of the
terms in the sequence are greater than 10?
(1) a1 = 24
(2) a8 = 10

OA:B

I think answer is actually C, but before explain my view, I'd like to know your own opinion.

Given a sequence: $$a_1, \ a_2, \ a_3, \ ... \ a_{14}, \ a_{15}$$

In the sequence shown, $$a_n = a_{n-1}+k$$, where $$2\leq{n}\leq{15}$$ and $$k$$ is a nonzero constant. How many of the terms in the sequence are greater than 10?

(1) $$a_1= 24$$
(2) $$a_8= 10$$

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 the value of $$k$$, 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.

Hope it helps.

P.S. Please do not reword the questions when posting (you omitted the crucial part: the sequence itself).

Bunuel,
I see you answering all questions beautifully.
I just want to know if you were able to get the solution within 2 minutes.

However, Thanks very much.
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11 Sep 2013, 06:44
1
SUNGMAT710 wrote:
Bunuel wrote:
mushyyy wrote:
In the sequence shown, a(n) = a(n-1)+k, where 2<=n<=15 and k is a nonzero constant. How many of the
terms in the sequence are greater than 10?
(1) a1 = 24
(2) a8 = 10

OA:B

I think answer is actually C, but before explain my view, I'd like to know your own opinion.

Given a sequence: $$a_1, \ a_2, \ a_3, \ ... \ a_{14}, \ a_{15}$$

In the sequence shown, $$a_n = a_{n-1}+k$$, where $$2\leq{n}\leq{15}$$ and $$k$$ is a nonzero constant. How many of the terms in the sequence are greater than 10?

(1) $$a_1= 24$$
(2) $$a_8= 10$$

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 the value of $$k$$, 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.

Hope it helps.

P.S. Please do not reword the questions when posting (you omitted the crucial part: the sequence itself).

Bunuel,
I see you answering all questions beautifully.
I just want to know if you were able to get the solution within 2 minutes.

However, Thanks very much.

Actually, writing this solution took much longer than solving itself. So, yes I think it's possible to solve the question under 2 minutes.
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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03 Nov 2013, 02:59
statement 2 is very very tricky great explanation as always Bunuel! Just added numerical examples for slow learners like me!

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 $$a_8$$.

Let K = -3

a1 to a7 => 31,28,22,19,16,13
a8 = 10
a9 to a15 = > 7,4,1,-2,-5,-8,-11

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 $$a_8$$;

Plug in examples here Lets say K = 2 so the sequence from
a1 to a7 => -4,-2,-0,2,4,6,8
a8 = 10
a9 to a15 => 12,14,16,18,20,22,24

Are there any similar questions to practice?
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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09 Jul 2015, 15:22
Bunuel: How would you rate the difficulty level of this question?
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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10 Jul 2015, 01:49
avgroh wrote:
Bunuel: How would you rate the difficulty level of this question?

~650 but users rate it much higher (harder).
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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03 Oct 2015, 05:20
Alright, maybe it's too late where i am but I must be missing something:

Isn't there 14 terms and not 15?

2<=n<=15
so n-->2,3,4,5,6,7,8,9,10,11,12,13,14,15

also 15-2+1 = 14
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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03 Oct 2015, 06:32
dor1209 wrote:
Alright, maybe it's too late where i am but I must be missing something:

Isn't there 14 terms and not 15?

2<=n<=15
so n-->2,3,4,5,6,7,8,9,10,11,12,13,14,15

also 15-2+1 = 14

The stem gives the sequence of 15 terms: Given a sequence: $$a_1, \ a_2, \ a_3, \ ... \ a_{14}, \ a_{15}$$.

The formula ($$a_n = a_{n-1}+k$$) simply defines terms from 2nd to 15th terms.
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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12 Apr 2016, 13:45
I agree with everything said above except one thing --- how can we assume value of a1? Minimum value n can take is 2 and series is defined a2=a1+k. for all i know a1 can be 100 or -100 and every word of the question will still be true.
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In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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26 Jun 2016, 17:43
definitely an easy problem to overthink. Thanks Bunuel!
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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26 Oct 2016, 12:28
1
This is definitely a C-Trap question. It is pretty tricky to catch the "median" and "sequence" point when the question just pops out in the middle of your test. +1 to Bunuel for explaining it so nicely.

Thank You.
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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01 Oct 2017, 13:02
Bunuel wrote:
mushyyy wrote:
In the sequence shown, a(n) = a(n-1)+k, where 2<=n<=15 and k is a nonzero constant. How many of the
terms in the sequence are greater than 10?
(1) a1 = 24
(2) a8 = 10

OA:B

I think answer is actually C, but before explain my view, I'd like to know your own opinion.

Given a sequence: $$a_1, \ a_2, \ a_3, \ ... \ a_{14}, \ a_{15}$$

In the sequence shown, $$a_n = a_{n-1}+k$$, where $$2\leq{n}\leq{15}$$ and $$k$$ is a nonzero constant. How many of the terms in the sequence are greater than 10?

(1) $$a_1= 24$$
(2) $$a_8= 10$$

We have a sequence of fifteen terms (actually this sequence is an 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 a 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 the value of $$k$$, 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.

Hope it helps.

P.S. Please do not reword the questions when posting (you omitted the crucial part: the sequence itself).

Hi Bunuel -- i see your solution but when i did this i thought the median was the average of a_8 and a_9

Why consider a_1 as part of your arithmetic progression when a_1 is not possible

I see in S1, a_1 is given

But as a test taker, how should a test taker think about including a_1

Please explain the 'logic' behind including a_1 in your progression (only this way can a_8 be the median)
Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k &nbs [#permalink] 01 Oct 2017, 13:02

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