A set consists of n distinct integers arranged in the order of increas

IMO B

Statement 1: The sum of any 3 successive integers of the set is divisible by 3

Case 1: 1,3,5,7,9
=> 3+5+7/3 => 15/3 = 5
=> Mean = 5 = Median (sufficient)

Case 2: 1,9,89,91,120
=> 9+89+91/3 = 189/3 = 63
=> Mean = 61.8(309/5) ≠ Median = 89 (Not sufficient)
Statement 1 alone is not sufficient.

Statement 2: The difference between any 2 successive integers of the set is 4 (since the difference between two no.is 4 the set become consecutive, and the mean is always equals to median in consecutive sets)

Case 1: 1,5,9,13,17
=> Mean = 9 (1+5+9+13+17/5)
=> 45/5 = 9
=> Median = 9 (sufficient)

Case 2: 2,6,10,14,18
=> Mean = 10 (2+6+10+14+18/5)
=> 50/5 = 10
=> median = 10 (sufficient)

Hence, statement 2 is sufficient

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statement 1:
any three successive integers are consecutive or all the numbers are multiples of 3. either way sequence will be in AP.
median = mean

statement 2:
common difference = 4
sequence is in AP then median = mean

Ans: D

IMO B

A set consists of n distinct integers arranged in the order of increasing magnitude. Is the median of the n integers equal to the arithmetic mean of the n integers:

(1) The sum of any 3 successive integers of the set is divisible by 3

Soln: Let n = { 1, 2, 3, 7 }
Mean = 13/4 & Median = 5/2 [Not equal]
n= {1, 2, 3}
Mean = 2 & Median = 2 [Equal]

Not Sufficient

(2) The difference between any 2 successive integers of the set is 4

Soln: n = { x, x+4, x+8, x+12 }
Mean = (4x+24)/4 = x+6
Median = (2x+12)/2 = x+6
Mean = Median

Sufficient

Statement 1 is not sufficient as it could be any random numbers hence mean and median may or may not be same
E.g:just take three numbers,
3,6,9, (same)
1,3,8. (not same)
Statement 2 is sufficient as, numbers are in arithmetic progression (constant difference), so mean is equal to median.

Ans B

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set n distinct integers
mean=median if set is in an arithmetic progression (AP)
ie. {a+d,a+2d,a+3d…etc}={2,4,6,8…}={10,40,70…}

(1) insufic
set {3,6,9} then mean=median
set {3,6,9,30} then mean≠median

(2) sufic
any two "successive" has dif of 4
set {4,8,12,…} this is an AP
set {10,14,18…} this is also an AP

Ans (B)

median and mean of set is equal only when set is evenly spaced
#1
The sum of any 3 successive integers of the set is divisible by 3
1,5,9 ; yes
1,3,8 ; no
insufficient
#2
The difference between any 2 successive integers of the set is 4
2,6,10,14 ; yes
3,7,11 ; yes
sufficient because median and mean of set is equal only when set is evenly spaced
OPTION B

A set consists of n distinct integers arranged in the order of increasing magnitude. Is the median of the n integers equal to the arithmetic mean of the n integers?

(1) The sum of any 3 successive integers of the set is divisible by 3
(2) The difference between any 2 successive integers of the set is 4

Is it B ?

A set consists of n distinct integers arranged in the order of increasing magnitude. Is the median of the n integers equal to the arithmetic mean of the n integers?

so set can be (0 , -1 , 6 ...) .
Is mean = Median ?
St (1) :
The sum of any 3 successive integers of the set is divisible by 3

If S = (0,1,2) mean = median
if S = (0,1,101) .. Mean <> median ...
so st 1 is insufficient .


St (2) :
The difference between any 2 successive integers of the set is 4 .

The set is in AP with common diference 4 .

S = (0, 4 , 8 ) Mean = Median = 4 (odd no of terms in an AP .. must be mean = median)

S = (12,16,20,24) Mean = Median = 18 .. (For even no of term , since the difference is divisible by 4 mean will be equal to median)

St 2 is sufficient .
So B .

A set consists of n distinct integers arranged in the order of increasing magnitude. Is the median of the n integers equal to the arithmetic mean of the n integers?

(1) The sum of any 3 successive integers of the set is divisible by 3..............for ex: 3,9,15.....which satisfies this condition and are part of this set but its mean and median are different..............INSUFFICIENT
(2) The difference between any 2 successive integers of the set is 4..............This says that they are in arithmetic progression.........so its mean and median will be same........SUFFICIENT

OA:B

A set consists of n distinct integers arranged in the order of increasing magnitude. Is the median of the n integers equal to the arithmetic mean of the n integers?

(1) The sum of any 3 successive integers of the set is divisible by 3
(2) The difference between any 2 successive integers of the set is 4

set ={a,b,c,...till n numbers}

1) a+b+c = 3k,
can be 1,3,5
can be 3,6,9
insufficient

2)b-a=4
differnece is same..so AP
sufficient

Ans B

A set consists of n distinct integers arranged in the order of increasing magnitude. Is the median of the n integers equal to the arithmetic mean of the n integers?

(1) The sum of any 3 successive integers of the set is divisible by 3
There are 2 possibilities:
1. Consecutive integers - 1,2,3,4 or 5,6,7,8,9. In both cases the mean and median are same
2. All are multiples of 3- 3,6,9,12 or 9,15,18,21 -> In the second case the mean and median is not same.

So insufficient.


(2) The difference between any 2 successive integers of the set is 4

1. 4,8,12,16 GOOD TO GO
2. 2,6,10,14 GOOD TO GO
Hence Sufficient

B