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The nth term, where n > 1, in a sequence S of positive integers is def

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

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New post 19 Nov 2019, 01:39
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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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Re: The nth term, where n > 1, in a sequence S of positive integers is def  [#permalink]

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New post 19 Nov 2019, 02:17
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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}

:please Pls give kudos, if you find my explanation good enough. :please
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The nth term, where n > 1, in a sequence S of positive integers is def  [#permalink]

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New post 19 Nov 2019, 07:05
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

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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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New post 19 Nov 2019, 07:39
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
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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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New post 19 Nov 2019, 07:45
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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New post 21 Nov 2019, 06:21
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.

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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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