For any positive integer n, the sum of the first n positive multiples

Question

For any positive integer n, the sum of the first n positive multiples of 3 equals 3n(n + 1)/2. What is the sum of all the multiples of 3 between 299 and 601 ?

A. 15,150
B. 22,275
C. 45,000
D. 45,450
E. 90,900

ScottTargetTestPrep · Accepted Answer

We can use the formula sum = average x quantity

Since we have an evenly spaced set, the average is:

(lowest multiple of 3 + highest multiple of 3)/2

(300 + 600)/2 = 900/2 = 450

Since we have an evenly spaced set the quantity is:

(highest multiple of 3 - lowest multiple of 3)/3 + 1

(600 - 300)/3 + 1= 100 + 1 = 101

Thus the sum is 450 x 101 = 45,450

Answer: D

bhima22 · Answer

For any positive integer n, the sum of the first n positive multiples of 3 equals 3n(n + 1)/2. What is the sum of all the multiples of 3 between 299 and 601 ?

A. 15,150
B. 22,275
C. 45,000
D. 45,450
E. 90,900

Answer: multiples of 3 between 299 and 601 = multiples of 3 from 100 - 200

Sum of multiples of 3 from 100 - 200 = Sum of the first 200 positive multiples - sum of the first 99 multiples

Sum of the first 200 positive multiples = 600 (201)/2 -> 60,300
Sum of the first 99 multiples = 297 (100)/2 --> 14,850

Substract this both to get -> 45,450  (D)

Solenja · Answer

Hey bhima22
Im completely lost could you please break it down a little more.
Thanks

bhima22 · Answer

Hi  Solenja .

- The formula given (3n(n + 1)/2) in the questions is for the sum of the FIRST n positive multiples of 3
- However, we're asked to find the sum of the multiples of 3 between 299 and 601 (therefore, we can't use plainly the formula given). 
- So, how can we use the formula given to answer the questions?

1) First can find the first multiples of 3 from 3 - 600 (This is the first 200 multiples of 3)
2) Then we find the first multiples of 3 from 3 - 297 (This is the first 99 multiples of 3)
3) Then, we can subtract the result from (1) and (2) to come up with the multiples of 3 between 299 - 601.

Hope it makes sense.

YannCJ · Answer

Please can someone help me understand the answer? thanks in advance!

purplex · Answer

Why can't I use the AP formula to find the sum of all multiples of 3 between 300 and 600, inclusive, using 300 as the first number and the difference as 3. Am I missing something with this approach?

youcouldsailaway · Answer

I think the formula given in the question stem to calculate the multiples of 3 is just given to confuse us.

We know that the first 100 multiples of 3 starts from 3  (3*1)  and end at 300  (3*100) . We know the first 200 multiples of 3 starts from 3 and end at 600  (3*200) .

The question wants us to calculate the sum of the multiple of 3 that fall between 299 and 601. So we can first calculate the sum of the first 99 multiples of 3 (since the question doesn't want anything less than 300) and then subtract them from the sum of the first 200 multiples of 3 using the  S_n  formula of AP.

Answer is D.

shubhim20 · Answer

why did we calculated 450 ? why not 3*101

Bunuel · Answer

The reason we calculate 450 is because it’s the average of the first and last multiples of 3 in the range:

(300 + 600) / 2 = 450

We use this average because the set of multiples of 3 from 300 to 600 is evenly spaced, so  the average times the number of terms gives the sum .

For any positive integer n, the sum of the first n positive multiples

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