Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1.
A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula.

In the even integers questions, you may be required to find the sum of first 10 even integers.
2 + 4 + 6 + ... + 18 + 20
Take 2 common, 2*(1 + 2 + 3 + ...10)
To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum.

Note that for odd integers, you cannot directly use this formula.
Sum the first 10 odd integers
1 + 3 + 5 + 7+...+19
But you can still make some modifications to find the sum.

1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) - (2 + 4+ 6+...20)
We know how to sum consecutive integers.
(1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2
(2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before)

So 1 + 3 + 5 + 7+...+19 = (20*21/2) - (10*11) = 100

The direct formula of sum of n consecutive odd integers starting from 1 = n^2
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3+6+9+12+15+18+21+24+27+30...... = 3*(1+2+3+4+5+6+7+8+9+10...)

So, yup.. You can do that...

Great explanation. However, I have a suggestion. Why can't we use one single formula for all consecutive evenly spaced numbers which is Sum = (average)(number of terms). I hope i am correct?
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We can, provided we know the average and the number of terms.
If we are asked to find the sum of first 50 consecutive positive odd integers, it might be easier to use 50^2 than to find average and then find the sum.

Mind you, I myself believe in knowing just the main all-applicable kind of formulas and then twisting them around to apply to any situation. But some people prefer to work more on specific formulas and these discussions are for their benefit.
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The best way to deal with sum of any type of arithmetic series is to use the following expression:

Sum = Average*Number of terms

Average = (First term + Last term)/2

Note: The average or the median can be related to the first and last terms, which are often given for these types of problems.

Number of Terms = (Last Term - First Term)/Spacing + 1

Also, I recommend that you understand the basis of these relationships as opposed to blindly memorizing them.

The above relationships are flexible enough to allow you to deal with any GMAT problem on these concepts.

For example in Q172 from the Official Guide: What is the sum of all the even integers between 99 and 301 ?

Average = (100+300)/2 = 200

Number of Terms = (300-100)/2 + 1 = 101 (Note spacing is 2 between consecutive even integers)

Sum = 200*101=20200

Cheers,
Dabral

for consecutive odd or even ranges

sum = (f + n-1) n

f is the first number in the sequence, n is the number of elements in the sequence

Hi mathsnoob,

You are correct that the formula that you listed can be used for adding consecutive even and odd integers. In fact, it can be derived from the general approach. For example, n would be the number of terms in the sequence. The last term would be equal to f + 2(n-1), the common difference here is 2 for consecutive even and odd integers. This means the average of the first and last term is equal to [f + f + 2(n-1)]/2, which simplifies to f + (n-1). And finally we have an expression for the sum of n consecutive integers n[f + n - 1].

The problem with these highly specialized formulas is the burden of remembering the specific formula and keeping track of what each term stands for. Personally, I prefer to stick to the general formula that I listed earlier and then adapt it to a specific situation. This helps reduce the memorization burden on my brain. But I can see others who might prefer to memorize formulas for specific situations. Take your pick.

Cheers,
Dabral

For more check Formulas for Consecutive, Even, Odd Integers

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