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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # The nth term, where n > 1, in a sequence S of positive integers is def

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Math Expert V
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The nth term, where n > 1, in a sequence S of positive integers is def  [#permalink]

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Question Stats: 52% (01:32) correct 48% (02:01) wrong based on 31 sessions

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The $$n_{th}$$ term, where n > 1, in a sequence S of positive integers is defined by $$t_n=t_{n-­1}+4$$. What is the standard deviation of the elements in S?

(1) The sequence contains 11 elements.
(2) The first element in the sequence is 7.

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Concentration: Finance, Operations
Re: The nth term, where n > 1, in a sequence S of positive integers is def  [#permalink]

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Bunuel wrote:
The $$n_{th}$$ term, where n > 1, in a sequence S of positive integers is defined by $$t_n=t_{n-­1}+4$$. What is the standard deviation of the elements in S?

(1) The sequence contains 11 elements.
(2) The first element in the sequence is 7.

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Answer is C , Both are required,
As by statement one alone, We cannot find all the terms of sequence which is required for standard deviation.

by statement 2, We can get all the terms, so easily find the standard deviation. But alone, we don't know what are the total numbers of terms.
S={7,11,14,19,23,27,31,35,39,44,47} Pls give kudos, if you find my explanation good enough. Math Expert V
Joined: 02 Aug 2009
Posts: 8336
The nth term, where n > 1, in a sequence S of positive integers is def  [#permalink]

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The $$n_{th}$$ term, where n > 1, in a sequence S of positive integers is defined by $$t_n=t_{n-­1}+4$$. What is the standard deviation of the elements in S?
So it is an AP with a difference of 4 between consecutive terms..
SD is dependent on the spread of numbers from MEDIAN or MEAN in an AP as Median=Mean in an AP. Here we know the spread of consecutive terms, so if we know the total numbers in the set, we would know the SD

(1) The sequence contains 11 elements.
EXACTLY what we were looking for. Irrespective of where the numbers lie, the SD will be same..
The spread will be Median-the extreme numbers, which is 4*4, and so on till the immediate numbers, which is 4, so 2((4*4)+(4*3)+(4*2)+(4*1)). This will be constant irrespective of set starting with 10, 100, 10000 etc
Suff

(2) The first element in the sequence is 7.
Insuff

A
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Re: The nth term, where n > 1, in a sequence S of positive integers is def  [#permalink]

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SD is dependent on the spread of numbers from MEDIAN.

I think it must be mean, spread from mean. Here given sequence is in AP, so MEAN= MEDIAN.
Am I correct?

chetan2u wrote:
he $$n_{th}$$ term, where n > 1, in a sequence S of positive integers is defined by $$t_n=t_{n-­1}+4$$. What is the standard deviation of the elements in S?
So it is an AP with a difference of 4 between consecutive terms..
SD is dependent on the spread of numbers from MEDIAN. Here we know the spread of consecutive terms, so if we know the total numbers in the set, we would know the SD

(1) The sequence contains 11 elements.
EXACTLY what we were looking for. Irrespective of where the numbers lie, the SD will be same..
The spread will be Median-the extreme numbers, which is 4*4, and so on till the immediate numbers, which is 4, so 2((4*4)+(4*3)+(4*2)+(4*1)). This will be constant irrespective of set starting with 10, 100, 10000 etc
Suff

(2) The first element in the sequence is 7.
Insuff

A
Math Expert V
Joined: 02 Aug 2009
Posts: 8336
Re: The nth term, where n > 1, in a sequence S of positive integers is def  [#permalink]

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gvij2017 wrote:
SD is dependent on the spread of numbers from MEDIAN.

I think it must be mean, spread from mean. Here given sequence is in AP, so MEAN= MEDIAN.
Am I correct?

chetan2u wrote:
he $$n_{th}$$ term, where n > 1, in a sequence S of positive integers is defined by $$t_n=t_{n-­1}+4$$. What is the standard deviation of the elements in S?
So it is an AP with a difference of 4 between consecutive terms..
SD is dependent on the spread of numbers from MEDIAN. Here we know the spread of consecutive terms, so if we know the total numbers in the set, we would know the SD

(1) The sequence contains 11 elements.
EXACTLY what we were looking for. Irrespective of where the numbers lie, the SD will be same..
The spread will be Median-the extreme numbers, which is 4*4, and so on till the immediate numbers, which is 4, so 2((4*4)+(4*3)+(4*2)+(4*1)). This will be constant irrespective of set starting with 10, 100, 10000 etc
Suff

(2) The first element in the sequence is 7.
Insuff

A

Yes, you are correct.
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The nth term, where n > 1, in a sequence S of positive integers is def  [#permalink]

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Bunuel wrote:
The $$n_{th}$$ term, where n > 1, in a sequence S of positive integers is defined by $$t_n=t_{n-­1}+4$$. What is the standard deviation of the elements in S?

(1) The sequence contains 11 elements.
(2) The first element in the sequence is 7.

$$sd=\sqrt{\frac{((a_1-mean)^2+(a_2-mean)^2…)}{n}}$$

set: {1,2,3} n: 3 mean: 2
term-mean: {-1,0,1}
sum squared: {1+0+1}=2
sd=√2/3

set: {2,3,4} n: 3 mean: 3
term-mean: {-1,0,1}
sum squared: {1+0+1}=2
sd=√2/3

set: {1,2,3,4,5} n: 5 mean: 3
term-mean: {-2,-1,0,1,2}
sum squared: {4+1+0+1+4}=10
sd=√10/5=√2

Conclusion: number of terms in an AP affects the sd, but the value of the first element does not.

Ans (A) The nth term, where n > 1, in a sequence S of positive integers is def   [#permalink] 21 Nov 2019, 06:21
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