## Confused with calculating the sum of consecutive numbers

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**Confused with calculating the sum of consecutive numbers**
* Tue Aug 17, 2010 2:20 pm*
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I don't know what I'm doing wrong!!

Ok so the formula I have for finding the sum of consecutive numbers is

[n(n+1)]/2

I tested this formula out by using an easy range of numbers: 2 to 8

so the total number of numbers from 2 through 8 inclusive is: (8-2) + 1 = 7

So now I plug it into the formula: [(7(7+1)]/2 --> 56/2 = 28

so the sum of the integers from 2 to 8 inclusive is 28.....

HOWEVER

I checked this by adding 2+3+4+5+6+7+8 and that gives me 35.

What am I doing wrong????

* Tue Aug 17, 2010 2:42 pm*
**szy wrote:**I don't know what I'm doing wrong!!

Ok so the formula I have for finding the sum of consecutive numbers is

[n(n+1)]/2

I tested this formula out by using an easy range of numbers: 2 to 8

so the total number of numbers from 2 through 8 inclusive is: (8-2) + 1 = 7

So now I plug it into the formula: [(7(7+1)]/2 --> 56/2 = 28

so the sum of the integers from 2 to 8 inclusive is 28.....

HOWEVER

I checked this by adding 2+3+4+5+6+7+8 and that gives me 35.

What am I doing wrong????

The formula you mention allows you to calculate the sum 1+ 2 + 3 + ... + n. That is, it gives the sum of the integers from 1 up to n, inclusive. If you don't start at 1, you'd need to modify the calculation somehow. There are quite a few ways this can be done. For example, if you want to find the sum 2 + 3 + 4 + ... + 8, well, that's almost the same thing as the sum 1+2+3+...+8, except that we don't want to include the 1. So if you use your formula with n=8, then subtract 1, you'll get the right answer: [8(8+1)/2] - 1 = 36 - 1 = 35.

There's a more general method to sum any 'equally spaced' list which I and others have described a few times on this forum, and which should be discussed in any decent prep book; if no one else explains it here, I will when I have a bit more free time.

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* Tue Aug 17, 2010 4:45 pm*
here it is your answer.

The formula you are already using is the sumatory from n=1 to K= n that equals to n(n+1)/2

Your original sumatory didn't start 1 it started on 2.

You can still use your formula but remember it the answer will be the sum from 1 to 8 that equals 36 as you stated, then you must substract 1 that is the difference between the first number and the consecutive one, that give you the new total of 35.

* Tue Aug 17, 2010 5:54 pm*
thanks guys that makes sense -- i found the formulas that are easier to use

* Tue Aug 17, 2010 5:58 pm*
You guys are probably missing something...

if you consider Sn = n (A1 + An)/2

where n = number of terms , A1 = the first term of the sequence, An = the last term of the sequence,

you'll get the answer..

* Fri Aug 20, 2010 9:29 pm*
the formula for calculating the sum of n terms where a is the first term and the common difference "d" is given by

Sn = n/2 [ 2a + (n-1) d ] or n/2 [ sum of first term and the last term ]

in your example

Sn = 7/2 [ 2(2) + (7-1)* 1 ] = 7/2 * 10 = 35

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@Deb