The average (arithmetic mean) of the multiples of 6

Question

The average (arithmetic mean) of the multiples of 6 that are greater than 0 and less than 1,000 is

A. 499 
B. 500 
C. 501 
D. 502 
E. 503

Bunuel · Accepted Answer

Multiples of 6 represent arithmetic progression (aka evenly spaced set). In any evenly spaced set the  arithmetic mean (average) is equal to the median  and can be calculated by the formula  mean=median=\frac{a_1+a_n}{2} , where  a_1  is the first term and  a_n  is the last term.

First term is 6 and last term is 996 (the last even multiple of 3 below 1000). So  mean=\frac{6+996}{2}=501 .

Answer: C.

perseverant · Answer

Thank you! this is very helpful!

First term is 6 and last term is 996 ( the last even multiple of 3 below 1000 ). So  mean=\frac{6+996}{2}=501 . 
I assume you meant multiple of 6 instead of 3.

Bunuel · Answer

No. This is how I found the last multiple of 6 below 1000: it would be the last  EVEN multiple of 3  (thus multiple of 6) below 1000 .

KrishnakumarKA1 · Answer

First term is 6 and last term is 996
Sum of an AP is {n( first term + last term)}/2
Average = {n( first term + last term)}/2n 
= ( first term + last term)/2 
= (6 + 996)/2 
= 501. Option C

JeffTargetTestPrep · Answer

Since we have an evenly spaced set, we can use the following average formula:

average = (first term in the set + last term in the set)/2

The first multiple of 6 in the set is 6 and the last multiple of 6 is 996. Thus:

average = (6 + 996)/2 = 1002/2 = 501

Answer: C

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The average (arithmetic mean) of the multiples of 6

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The average (arithmetic mean) of the multiples of 6

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