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03 Nov 2005, 23:05
Kudos
Bookmarks
Hi Can anyone please give me a clarification on how to find the sum of consecutive odd/even numbers?
I read the thread below that
https://www.gmatclub.com/phpbb/viewtopi ... g+integers
However, if I apply the same concept to find the sum of even integers from 5 to 9, inclusive, it doesn't seem to work. I was only able to find the sum of odd integers using that formula.
Are there any variations?
Thanks.
I read the thread below that
https://www.gmatclub.com/phpbb/viewtopi ... g+integers
ian7777
However, if I apply the same concept to find the sum of even integers from 5 to 9, inclusive, it doesn't seem to work. I was only able to find the sum of odd integers using that formula.
Are there any variations?
Thanks.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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03 Nov 2005, 23:18
Kudos
Bookmarks
Well the generic way to sum a set of consecutive numbers is to follow our friend Gauss approach:
Avg of First and last term * number of items in the set
for example:
to add:
1 + 2 + 3 +.. + 100
(100 + 1) / 2 * 100
2 + 4 + 6 + ...+ 200 =
(200 + 2)/2 * 100
the quoted section in your post talks about finding "number of items in the set".
for eample, in the above sequence of even numbers, there are 100 items. How do you find it? The generic way is :
(last element - first element/interval) + 1
so, for even number series like above, the number of items in the series is:
(200 - 2 / 2) + 1 = 100
The think to remember is that you should apply this when first and last element is in the set.
for example, if you are to find sum of even numbers from 1 to 99, first find first even number and last even number:
first element: 2
last element: 98
now apply above technique
Avg of First and last term * number of items in the set
for example:
to add:
1 + 2 + 3 +.. + 100
(100 + 1) / 2 * 100
2 + 4 + 6 + ...+ 200 =
(200 + 2)/2 * 100
the quoted section in your post talks about finding "number of items in the set".
for eample, in the above sequence of even numbers, there are 100 items. How do you find it? The generic way is :
(last element - first element/interval) + 1
so, for even number series like above, the number of items in the series is:
(200 - 2 / 2) + 1 = 100
The think to remember is that you should apply this when first and last element is in the set.
for example, if you are to find sum of even numbers from 1 to 99, first find first even number and last even number:
first element: 2
last element: 98
now apply above technique
I understood now.
