# Traversing Averages

by on September 6th, 2012

In an earlier article we discussed that the average of a set of consecutive numbers equals the median of the set. Now let’s discuss another important property of consecutive integers.

Average of a list of consecutive integers = Average of any equidistant pair of integers around the median.

For example, take a set of consecutive integers {1, 2, 3, 4, 5}. ”3″ is the median, and therefore the average.

Take a look at the pair “2″ and “4″ (one step away from the average 3): 2 is one down from 3, 4 is one up from 3. Adding the two will cancel the difference and leave “6″. The average of 2 and 4 is 6/2 = 3 = the average of the group.
The same happens if we take another step away from the median – the average of 1 and 5 is still 3 which is the median and the average of the entire set.

The use of this insight depends on what data is provided by the question. In many cases, GMAT questions will describe a set of consecutive integers in terms of its first and last terms – all the integers between 22 and 55, for example. Since the average of the entire group is equal to the average of any pair of terms around the middle, we can use the first and last term provided by the above statement to calculate the average of the entire set.

In other words, the average of the set of all integers between 22 and 55 is simply the average of the first and last terms of the set: the average of 22 and 55 = (22+55) / 2 = 77 / 2 = 38.5. This will also be the median of the set, according to what we’ve already learned of the properties of the average of a set of consecutive integers.

Remember that this insight holds true for any set of integers with a constant difference between any two consecutive terms (arithmetic sequence) – not just for consecutive integers.

For example, the multiples of 3 between 22 and 55 constitute a set of integers with a constant difference of 3 between any two terms, and thus the properties of consecutive integers apply in this case as well. If the question asks about the average of a set of all multiples of 3 between 22 and 55 (inclusive), simply find the first and last terms of the set – the first and last multiples of three within that range. Averaging the two multiples will give you the average of the entire set:

22 isn’t a multiple of 3 – the first multiple of 3 within the specified range is 24.

55 isn’t a multiple of 3 either – the last multiple of 3 within the specified range is 54.

Therefore, the average of the set of multiples of 3 between 22 and 55 = the average of the first and last multiple of 3 within the range = (24+54) / 2 = 78 / 2 = 39.

## Remember:

For any set of integers with a constant difference between any two consecutive terms (such as consecutive integers or multiples):

Average of the set = Average of any equidistant pair of terms around the median.

From this insight, remember this simpler rule:

Average of a Set of consecutive integers = Average of the first and last terms of the set.

Now try solving the following question using the above method:

If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

• C

• C

• Deceptively hard question.

• Nice, loved it. The answer is C.

• nice one ........

• (108+495)/2=301.5

• How did you know to choose 108 and 495? I thought of 109 and 491 but, of course, get 300 and I'm sure that is the trick answer.

• Isn't C the average? I think E is the median as it is the number in the 23 place out of the 45 N set. Please let me know.... Cheers

• I don't know how C is the answer. I am getting E same as Yoav. I am simply applying arithmetic series formula: a+(n-1)*d.

A is 108; n will be 23 (remember median is 23:
(45+1)/2; d is 9.

Let me know if you agree.