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

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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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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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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k [#permalink]
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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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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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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k [#permalink]
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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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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k [#permalink]
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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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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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]
What is stopping A1 being -10 and K is -10. Then wouldn't it just oscillate between -10 and 10, and the answer is 0?
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k [#permalink]
rohit60 wrote:
What is stopping A1 being -10 and K is -10. Then wouldn't it just oscillate between -10 and 10, and the answer is 0?

In this case the sequence would be:

-10, -20, -30, -40, -50, -60, -70, -80, ... So, in this case, a8 would be -80 and not 10 as given in the second statement.
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k [#permalink]
Bunuel wrote:
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.

Here n could be any number right less than 15 and more than 8 then median could change if N =13then answer would be different , can anyone explain this
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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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SantoshN wrote:
Bunuel wrote:
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.

Here n could be any number right less than 15 and more than 8 then median could change if N =13then answer would be different , can anyone explain this

n there is an index number and shows which position a term holds in the sequence. For example, $$a_7$$ is the 7th term in the sequence. The given sequence has 15 terms so, n is from 1 to 15, inclusive.
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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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mushyyy wrote:
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) $$a_8= 10$$

I love this question!!

Given: In the sequence shown, $$a_n = a_{n-1}+k$$, where $$2\leq{n}\leq{15}$$ and $$k$$ is a nonzero constant.
IMPORTANT: This sequence has exactly 15 terms.
Also, since k doesn't equal 0, the sequence EITHER increases with each subsequent term OR decreases with each subsequent term.

Target question: How many of the terms in the sequence are greater than 10?

Statement 1: $$a_1= 24$$
This doesn't help us answer the target question.
Consider these two possible cases:
Case a: k = 2, which means each term is 2 greater than the previous term. So our sequence looks like this: 24, 26, 28, 30, etc, In this case, the answer to the target question is all 15 terms are greater than 10
Case b: k = -10, which means each term is 10 less than the previous term. So our sequence looks like this: 24, 14, 4, -6, -16, etc, In this case, the answer to the target question is 2 terms are greater than 10
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: $$a_8= 10$$
IMPORTANT: Notice that $$a_8$$ is the middle term among all 15 terms.
We already know that the sequence EITHER increases with each subsequent term OR decreases with each subsequent term. So let's consider each possible case:
Case a: The sequence INCREASES with each subsequent term. In this case, terms $$a_1$$ to $$a_7$$ will be less than 10, and terms $$a_9$$ to $$a_{15}$$ will be greater than 10. So, the answer to the target question is 7 terms are greater than 10
Case b: The sequence DECREASES with each subsequent term. In this case, terms $$a_1$$ to $$a_7$$ will be greater than 10, and terms $$a_9$$ to $$a_{15}$$ will be less than 10. So, the answer to the target question is 7 terms are greater than 10
Since both possible cases yield the SAME answer to the target question ( 7 terms are greater than 10), we can answer the target question with certainty.
Statement 2 is SUFFICIENT

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

I was thrown off by the fact that the problem said 2<= n <= 15, thinking that this meant that we were only interested in n2 onwards of the sequence--basically eliminating the first term and making the sequence 14 terms instead of 15. I realize now that I miss understood the notation. Can you please clarify if 2<= n <= 15 is part of standard notation for this kind of sequence?
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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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acai2y wrote:
Bunuel wrote:
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).

I was thrown off by the fact that the problem said 2<= n <= 15, thinking that this meant that we were only interested in n2 onwards of the sequence--basically eliminating the first term and making the sequence 14 terms instead of 15. I realize now that I miss understood the notation. Can you please clarify if 2<= n <= 15 is part of standard notation for this kind of sequence?

We are given an arithmetic progression, which consists of 15 terms. "$$a_n = a_{n-1}+k$$, where $$2\leq{n}\leq{15}$$ and $$k$$ is a nonzero constant", means that the formula to get the nth term ($$a_n = a_{n-1}+k$$) "works" for terms from $$a_2$$ to $$a_{15}$$. So:
$$a_2=a_1+k$$;
$$a_3=a_2+k$$;
...
$$a_{15}=a_{14}+k$$.

You cannot apply it to the first term, though.
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k [#permalink]
Bunuel wrote:
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).

7 terns less than 10, and 8 terms greater than 10 if k is positive
But, opposite when k is negative?
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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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lakshya14 wrote:
Bunuel wrote:
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).

7 terns less than 10, and 8 terms greater than 10 if k is positive
But, opposite when k is negative?

For (2) 7 terms will be more than $$a_8=10$$ and 7 terms will be less than $$a_8=10$$ in both cases. Check highlighted part.
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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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