Problem Solving

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

OA : C Source : Kaplan Online

Just to add more:
In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula mean=median=(a_1+a_n)/2, where a_1 is the first term and a_n is the last term

since 6 is divisible by 3, just look for a number that is divisible by 3... which in this case is 501

komal wrote: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

OA : C Source : Kaplan Online

Some facts we should know about any set of evenly spaced, consecutive integers:

1. Average = median = (biggest + smallest)/2
{1, 2, 3, 4, 5, 6, 7, 8, 9}
Average = median = (9+1)/2 = 5.

2. Number of integers = (biggest - smallest)/(difference between each successive pair of integers) + 1
{1, 2, 3, 4, 5, 6, 7, 8, 9}
Difference between each successive pair = 1.
Number of integers = (9-1)/1 + 1 = 9.

{2, 4, 6, 8}
Difference between each successive pair = 2
Number of integers = (8-2)/2 + 1 = 4

{3, 6, 9}
Difference between each successive pair = 3
Number of integers = (9-3)/3 + 1 = 3.

3. Sum = (average) * (number of integers)
{1, 2, 3, 4, 5, 6, 7, 8, 9}
Average = (9+1)/2 = 5
Number of integers = (9-1)/1 + 1 = 9.
Sum = 5*9 = 45

{2, 4, 6, 8}
Average = (8+2)/2 = 5.
Number of integers = (8-2)/2 + 1 = 4
Sum =5*4 = 20.

{3, 6, 9}
Average = (9+3)/2 = 6.
Number of integers = (9-3)/3 + 1 = 3.
Sum = 6*3 = 18.

A harder example:
What is the sum of all the multiples of 7 between 1 and 100?
Range is 7 to 98.
Average = (98+7)/2. (This is also the value of the median.)
Number = (98-7)/7 + 1.
Sum = average * number = (98+7)/2 * ((98-7)/7 + 1) = (105/2)*14 = 735.

Hope this helps!