What is the sum of all multiples of 3 between 1 and 200?

Max@Math Revolution wrote: What is the sum of all multiples of 3 between 1 and 200?

A. 3300
B. 6600
C. 6633
D. 10100
E. 20100

? = 3 + 6 + 9 + ... + 195 + 198

In any finite arithmetic progression, the (global) average is equal to the median and to the average of symmetric elements (to the median).

The first and last elements are always symmetrical, hence their average (3+198)/2 is equal to the (global) average.

How many terms are involved?

3⩽3M⩽198(=180+18) hence 1⩽M⩽66 , therefore 66 terms.

(In this case, the same 66 may be obtained by 198/3 , of course.)

Using the homogeneity nature of the average we have:

? = 66⋅(3+198)/2 = 33⋅201 , therefore ⟨?⟩ = ⟨33⋅201⟩ = ⟨3⟩ where ⟨N⟩ = unit´s digit of N

The above follows the notations and rationale taught in the GMATH method.

=>

We need to find the sum of 3, 6, …, 198.
The number of data values is (198 – 3)/3 + 1 = 195/3 + 1 = 65 + 1 = 66.
Thus the sum of 3, 6, …, 198 is 3 + 6 + … + 198 = 3+6+…+198=3(1+2+…+66)=3(66)(66+1)/2=3(33)(67)=6,633.

Therefore, the answer is C.
Answer: C

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

What is the sum of all multiples of 3 between 1 and 200?

A. 3300
B. 6600
C. 6633
D. 10100
E. 20100

The smallest multiple of 3 between 1 and 200 is 3, and the greatest multiple is 198. Thus, the number of multiples of 3 between 1 and 200 is (198 - 3)/3 + 1 = 65 + 1 = 66.

Now we use the formula for the average of the members of an evenly spaced set. The average of the 66 multiples of 3 is (smallest + greatest)/2 = (3 + 198)/2 = 201/2.

Since sum = average x quantity, the sum = 201/2 x 66 = 201 x 33 = 6633.

Answer: C