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If the average (arithmetic mean) of the first sixteen positive integer

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If the average (arithmetic mean) of the first sixteen positive integer [#permalink] New post   20 Jun 2020, 01:13
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If the average (arithmetic mean) of the first sixteen positive integers is subtracted from the average (arithmetic mean) of the next sixteen positive integers, what is the result?

(A) 0
(B) 16
(C) 32
(D) 64
(E) 128
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If the average (arithmetic mean) of the first sixteen positive integer [#permalink] New post   20 Jun 2020, 01:17
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SajjadAhmad wrote:
If the average (arithmetic mean) of the first sixteen positive integers is subtracted from the average (arithmetic mean) of the next sixteen positive integers, what is the result?

(A) 0
(B) 16
(C) 32
(D) 64
(E) 128


CONCEPT: FOr the terms in Arithmetic Progression, \(Mean = Median = \frac{First+Last}{2}\)


Method 1:



Average of First sixteen positive integers = Median = (8th Term + 9th Term)/2 = 8+9/2 = 8.5

Average of Next sixteen positive integers (17 to 32) = (1st Term + Last Term)/2 = (17+32)/2 = 24.5

Required Difference = 24.5 - 8.5 = 16

Answer: Option B

Method 2:



Every term of next 16 terms is 16 greater than every respective term of first 16 terms

i.e. The Mean of next 16 terms will also be 16 greater than mean of first 16 terms

Answer: Option B
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Re: If the average (arithmetic mean) of the first sixteen positive integer [#permalink] New post   20 Jun 2020, 01:56
32 positive number = (N),(N+1),(N+2),(N+3),(N+4),(N+5),(N+6),(N+7),(N+8),............(N+31),(N+32)

Sum of first 16 positive number = (N)+(N+1)+(N+2)+(N+3)+(N+4)+(N+5)+(N+6)+(N+7)+(N+8)+(N+9)+(N+10)+(N+11)+(N+12)+(N+13)+(N+14)+(N+15) => 16N+ 120

Average of first 16 positive number= (16N+120)/16

=> Sum of Next 16 positive number = (N+16)+(N+17)+(N+18)+(N+19)+(N+20)+(N+21)+(N+22)+(N+23)+(N+24)+(N+25)+(N+26)+(N+27)+(N+28)+(N+29)+(N+30)+(N+31) => 16N+376

Average of first 16 positive number= (16N+376)/16

Difference between the first sixteen positive integers and the next sixteen positive integers =
=> 16N+376-16N-120/16
=> 256/16 => 16

Answer is B

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Re: If the average (arithmetic mean) of the first sixteen positive integer [#permalink] New post   20 Jun 2020, 09:10
SajjadAhmad wrote:
If the average (arithmetic mean) of the first sixteen positive integers is subtracted from the average (arithmetic mean) of the next sixteen positive integers, what is the result?

(A) 0
(B) 16
(C) 32
(D) 64
(E) 128


sum of 1-16th term ; 16*17/2 ; 136
and sum of 16th to 32nd term ; 32*33/2 - 16/17*2 ; 392
so avg ∆ 392/16-136/16 ; 256/16 ; 16
OPTION B
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Re: If the average (arithmetic mean) of the first sixteen positive integer [#permalink]
20 Jun 2020, 09:10
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