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A set consists of n positive integers arranged in decreasing order ...

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A set consists of n positive integers arranged in decreasing order ...  [#permalink]

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New post 09 Jan 2019, 02:22
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A set consists of n positive integers arranged in decreasing order of value. If the difference between any two adjacent numbers of the set is 3, then what is the ratio of arithmetic mean of the set to the median of the set?

    A. 1
    B. 1.5
    C. 2
    D. 2.5
    E. 3

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Re: A set consists of n positive integers arranged in decreasing order ...  [#permalink]

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New post 09 Jan 2019, 02:38
test with both odd & even set of n integers in set n
n= ( 12,9,6,3) and n= (12,9,6)

for n even integers
mean = 30/4 = 7.5 and median 6+9/2 = 7.5
ratio = 7.5/7.5 = 1

for n odd integers
mean = 27/3 = 9 and median 9
ratio = 9/9 = 1
IMO A ; 1



EgmatQuantExpert wrote:
A set consists of n positive integers arranged in decreasing order of value. If the difference between any two adjacent numbers of the set is 3, then what is the ratio of arithmetic mean of the set to the median of the set?

    A. 1
    B. 1.5
    C. 2
    D. 2.5
    E. 3

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Re: A set consists of n positive integers arranged in decreasing order ...  [#permalink]

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New post 09 Jan 2019, 16:39
My approach was that in an arithmetic progression median = mean.

Hence A.

Or, Am I wrong?

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Re: A set consists of n positive integers arranged in decreasing order ...  [#permalink]

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New post 11 Jan 2019, 07:10

Solution


Given:
    • A set of n positive integers arranged in decreasing order of value
    • Difference between any two consecutive numbers of the sequence = 3

To find:
    • The ratio of arithmetic mean of the set to the median value of the set

Approach and Working:
    • We know that the median of the set depends upon the number of terms in the set.

Case 1: n is odd
The terms of the set can be written as {a, a – 3, a – 3*2, a – 3*3, …. , a – 3(n-1)}
    • Median = the middle term = \((n + \frac{1}{2})^{th} term = a – 3[\frac{(n + 1)}{2} – 1] = a – \frac{3(n – 1)}{2}\)
    • Mean = (first term + last term)/2 = \(a + a – \frac{3(n – 1)}{2} = a – \frac{3(n – 1)}{2}\)
    • Thus, median = mean

Case 2: n is even
The terms of the set can be written as {a, a – 3, a – 3*2, a – 3*3, …. , a – 3(n-1)}
    • Median = average of the middle two terms = \(\frac{1}{2} * [(\frac{n}{2})^{th} term + (\frac{n}{2} + 1)^{th} term] = \frac{1}{2} * [a – 3[(\frac{n}{2} – 1] + a – 3(\frac{n}{2} + 1 – 1)] = \frac{(2a - 3n + 3)}{2} = a – \frac{3(n – 1)}{2}\)
    • Mean = (first term + last term)/2 = \(a + a – \frac{3(n – 1)}{2} = a – \frac{3(n – 1)}{2}\)
    • Thus, median = mean

Therefore, for any value of n, the ration of median to mean of the set is 1

Hence, the correct answer is option A.

Answer: A

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Re: A set consists of n positive integers arranged in decreasing order ...   [#permalink] 11 Jan 2019, 07:10
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