Q1, Q2, Q3, …, Qn is a sequence of n consecutive multiples of positive

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Q1, Q2, Q3, …, Qn is a sequence of n consecutive multiples of positive integer k. If n > 4, then what is the average (arithmetic mean) of the sequence?

(1) k = 4 
(2) The sum of the second greatest and the second least terms of the series is 24.

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yashikaaggarwal · Answer

(1) k = 4
So the sequence will be
4(1), 4(2), 4(3), 4(4),4(5).........
Average of set till 4(5) = 12
Average of set till 4(6) = 14
Average of set till 4(7) = 16
The average is varying.. 
Hence Insufficient.

(2) The sum of the second greatest and the second least terms of the series is 24
K(2) + K(n-1) = 24
K{2+n-1} = 24
K{n+1} = 24
Factors of 24 = 2*2*2*3
Since n > 4 
N+1 can only be 8 or 12
Then n = 7 or 11
If N = 11
K = 2
Sequence will be 2(1), 2(2).......2(11) 
And average 2(6) = 12

If N = 7
K = 3
Sequence will be 3(1), 3(2).......3(7) 
And average 3(4) = 12

(Sufficient)

IMO B

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unraveled · Answer

Q1, Q2, Q3, …, Qn is a sequence of n consecutive multiples of positive integer k. If n > 4, then what is the average (arithmetic mean) of the sequence?

Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}  = ?

(1) k = 4
Case1: Let  Q_1 = k, Q_2 = 2k, Q_3 = 3k ... Q_{10} = 10k 
 Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n} 
 Avg. = \frac{k + 2k + 3k + 4k + .... 10k}{10} 
 Avg. = \frac{11}{2}*k = 22

Case 2: Let  Q_1 = 5k, Q_2 = 6k, Q_3 = 7k ... Q_5 = 9k 
 Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n} 
 Avg. = \frac{5k + 6k + 7k + 8k + 9k}{5} 
Avg. = 7k = 28
 
INSUFFICIENT.

(2) The sum of the second greatest and the second least terms of the series is 24.
 Q_2 + Q_{n-1} = 24

Case 1: Let  Q_1 = k, Q_2 = 2k, Q_3 = 3k ... + Q_{10} = 10k 
2k + 9k = 24
 k = \frac{24}{11} 
 Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n} 
 Avg. = \frac{k + 2k + 2k + 4k + ..... + 10k}{10} 
 Avg. = \frac{11}{2}k = \frac{11}{2}*\frac{24}{11} = 12

Case 2: Let  Q_1 = 5k, Q_2 = 6k, Q_3 = 7k ... Q_5 = 9k 
6k + 8k = 24
 k = \frac{24}{14} 
 Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 + Q_5}{5} 
 Avg. = \frac{5k + 6k + 7k + 8k + 9k}{5} 
 Avg. = 7k = 7*\frac{24}{14} = 12

SUFFICIENT.

Answer B.

QuantMadeEasy · Answer

Step 1: Understanding the question 
When there is a sequence of consecutive multiples of a positive integer, the sequence is in Arithmetic Progression ie AP.
In an AP Mean or Average = Median ie. the middle term

Step 2: Understanding the statement(s) 
 
As the number of integers cannot be determined, the median or average of the list cannot be calculated
Insufficient

In an AP, Average or Median = first term + Last term/2 OR Second term + second last term/ 2 = 24/2 = 12
Sufficient

IMO (B)

minustark · Answer

Q1, Q2, Q3, …, Qn is a sequence of n consecutive multiples of positive integer k. If n > 4, then what is the average (arithmetic mean) of the sequence?

(1) k = 4
(2) The sum of the second greatest and the second least terms of the series is 24.

1) When, n = 5, the sequence becomes 4, 8, 12, 16, 20, and the average is 12. Again when n =6, the average of sequence 4, 8, 12, 16, 20, 24 will be 14. Not sufficient.

2) When k is 2, then it will be the sequence of first 12 positive multiples of 2, the average will be 12. Again when k =3, the sequence will be the first 7 positive multiples of 3. The average will be 12. It happens so as the average of equally spaced set can be expressed as the average of each pair consisting of two items; one from the rightmost and another from the leftmost, and so on. Example: 2, 4, 6, 8 ,10. Average is 6, which can be expressed as (2+10)/2 or, (4+8)/2. Sufficient.

B is the answer.

Lipun · Answer

The sequence is in AP.
Average of the sequence will be given by the average of - 
a) first term + last term
b) second term + second last term
... so on..

ST1: K=4. Insufficient

ST2: Sum of second term and second last term is 24. Average of sequence = 24/2 = 12. Sufficient.

Ans: B

brains · Answer

Q1, Q2, Q3, …, Qn is a sequence of n consecutive multiples of positive integer k. If n > 4, then what is the average (arithmetic mean) of the sequence?

(1) k = 4
(2) The sum of the second greatest and the second least terms of the series is 24.

Q Stem analysis:
For any integer , average of consecutive multiples of positive integer K will be in the form of : K*(n1 + n2 + .....+n)/n , where n is the multiplier such as 1,2 ,3 4 etc.

Statement 1: K=4 
We know the Integer which is K = 4 but we do not the value of n here. If n1 =1 , n2=2, n3=3, n4=4, n5=5 , then # of terms i.e. n= 5 , then average would be 12. But if n= 9 , then average would be 20. Hence insufficient:

Statement 2: As it is evenly spaced set, distance between the average and second greatest and second least would be same. Hence if 2nd greatest number is a and 2nd least number is b then a+b =24. And (a+b)/2 = 12 which will be the average of whole set. Hence statement 2 is sufficient.

Answer should be B

monikakumar · Answer

Q1, Q2, Q3, …, Qn is a sequence of n consecutive multiples of positive integer k. If n > 4, then what is the average (arithmetic mean) of the sequence?

(1) k = 4
(2) The sum of the second greatest and the second least terms of the series is 24.

sequence = k,2k,3k,4k,...(n-1)k,nk

1) average cannot be found without n
k=4
insufficient

2) 2k+(n-1)k=24
two variables cannot find average
insufficient

1+2)n can be found,
so average can be found
sufficient

Ans C

chetan2u · Answer

Q1, Q2, Q3, …, Qn is a sequence of n consecutive multiples of positive integer k. If n > 4, then what is the average (arithmetic mean) of the sequence?

(1) k = 4 
(2) The sum of the second greatest and the second least terms of the series is 24.

Consecutive multiples means it is an Arithmetic Progression 
An AP has its mean equal to its median.

So, we require to find the MEDIAN of sequence and not each item.

(1) k = 4 
We cannot find the median or the mean. If SUM is known, we can find the mean by dividing it by 4.
 
(2) The sum of the second greatest and the second least terms of the series is 24.
 Now, in an AP, the terms are evenly spread, so the MEDIAN can be found by taking average of the two terms equally placed from each end. for example largest and lowest or 2nd from top and below, etc. 
SO Median = mean = 24/2=12
Suff

B

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Q1, Q2, Q3, …, Qn is a sequence of n consecutive multiples of positive

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