How to - summing a list of consecutive numbers

[cbenk121]

Had a tricky problem in OG today (#116 if you're curious). Found the pattern, but then didn't calculate the sum correctly (ironically, you didn't even need to sum anything for that problem). But I could've solved it with a sum, had I calculated it correctly.

So then I set about to figure out how to, in general, calculate a list of consecutive numbers - if you happen to stumble on one. Maybe it's been posted before, but if you're like me and haven't seen it, here's a re-post then .

1) Find what the first and last term added together equal. This is the multiple.
2) Next, divide the last term by 2. This is how many pairs of the multiple you'll have.
2a) If the result is not an integer, then there will be an unpaired number in the middle. Round up to determine this middle number.
3) Multiply the multiple by the number of pairs, and add the middle number to the sum (if applicable).

For example: 1 to 59.
1 + 59 = 60. This is the multiple.
59 / 2 = 29.5. There will be 29 pairs of 60.
29.5 is not an integer, so rounding up, 30 is the middle number.
Our sum is 60 * 29 + 30 = 1770.

For example: 1 to 60
1 + 60 = 61. This is the multiple.
60 / 2 = 30. There will be 30 pairs of 60.
30 is an integer, so no middle number.
Our sum is 61 * 30 = 1830.

[GMATBootcamp]

Here's another common way to quickly calculate the sum of consecutive multiples or integers:

1. find the total number of integers in the set

(last number - first number)/multiple + 1 = total # integers in set

2. determine the average of the set

(last number + first number)/2 = average of set

3. multiply total # integers by the average to get the sum

total # integers * average of set = sum of consecutive numbers

eg.

What is the sum of the consecutive even integers from 2-60?

1. (60-2)/2 + 1 = 30
2. (60+2)/2 = 31
3. 30*31 = 930

[Testluv]

Hi cbenk121,

your method of studying for GMAT math is wonderful!

Unfortunately, your conclusion was just a little bit wrong in this case. Following your approach, what would the sum of 2+3+4 be?

First term + Last term = 6
Last term divided by two: 4/2 = 2
2*6= 12, but the sum of 2+3+4 is clearly 9.

Here is a safe way of doing it:
First, add the first and last terms.
Then, multiply by the number of terms.
Then, divide by two.

For the example of 2+3+4:
2+4 = 6
6*3 = 18
18/2 = 9

Also, you can use this technieque whenever the numbers are consecutive so long as the difference between any two terms is common; that is, so long as they are in an arithmetic (rather than a geometric) sequence. For example, what is the sum of 2+4+6 according to this approach?

2+6 = 8
8*3 = 24
24/2 = 12

If you're a formula person:

The sum of the terms in an arithmetic sequence=
n(first term + last term) divided by 2
where "n" is the number of terms in that sequence.

Oh, and to find the number of terms in a range of objects where the endpoints are included, you have to find the difference, and then add one. For example, the number of numbers in the range 10 to 60 inclusive is: 60-10 + 1= 51

[khushboo143]

thanks the above strategy did help me out...
i have one more doubt...i came across 1 question where the sum of consecutive odd nos. is asked.
can anyone help me out with this?

[Testluv]

Hi khushboo,

The above strategy would work. If it is concecutive odd or conesecutive even, the common difference between the terms is clearly two. If the set is too long to count the number of terms, then to count the number of terms in these (consecutive odd or consecutive even) sets:
Find the difference between the lowest and highest terms;
divide that by two;
then, add one.
Plug into the formula, and you have the sum.

We divide by two (step two) because the common difference is two. If the common difference was 4, say consecutive multiples of 4 (4+8+12+...), then, the common difference is four, and so we would divide by four (in order to count the number of terms in the set).

But there are other ways of solving these problems. Depending on the problem, simply picking numbers has a good chance of being a quicker method.