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27 Feb 2011, 02:09
Kudos
Bookmarks
Hey all,
I know i can find it in the forums but i failed.
If I have a set with a constant d (difference between the numbers) i have 2 ways to calculate the sum of the items:
(a1+an)*n/2
and
2/n*(2*a1 + (n-1)*d)
can someone plz tell me what is the difference? which one u use when and why?
thanks.
I know i can find it in the forums but i failed.
If I have a set with a constant d (difference between the numbers) i have 2 ways to calculate the sum of the items:
(a1+an)*n/2
and
2/n*(2*a1 + (n-1)*d)
can someone plz tell me what is the difference? which one u use when and why?
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
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
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27 Feb 2011, 02:34
Kudos
Bookmarks
144144
You are right;
If you know the first and the last term and number of elements; you use the first one;
First sequence;
2,...............................,16
Number of Elements, n = 8
Common difference, d=2 (not required)
\(Sum = \frac{(a_1+a_n)}{2}*n\)
\(Sum = \frac{(2+16)}{2}*8 = 72\)
If the sequence is;
2,4,6...
Number of elements, n=8
Common difference, d=2
Here you don't know the last term
So; you use the latter
\(Sum = \frac{n}{2}(2a_1+(n-1)d)\)
\(Sum = \frac{8}{2}(2*2+(8-1)2) = 72\)
As you can see, the first one is easily found. You use the first on all evenly spaced set where first and last term are known. If you don't know both first and last term, the latter formula should be used.
08 Mar 2011, 22:50
Kudos
Bookmarks
thank you fluke, you are really helpful!
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
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
