SOLUTION: Calculation of the following: 1+2+3+4....+97+98+99+100 I tried an arithmetic sequence a=1(100-1)1 but it wont work.

Algebra ->  Sequences-and-series -> SOLUTION: Calculation of the following: 1+2+3+4....+97+98+99+100 I tried an arithmetic sequence a=1(100-1)1 but it wont work.      Log On


   



Question 826123: Calculation of the following:
1+2+3+4....+97+98+99+100
I tried an arithmetic sequence a=1(100-1)1 but it wont work.

Found 3 solutions by richwmiller, stanbon, MathTherapy:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
100+50 +
1+99+2+98
4900+150=5050

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Calculation of the following:
1+2+3+4....+97+98+99+100
-------
s(n) = (n/2)(a + L)
----
s(100) = (100/2)(1+100)
----
s(100) = 50*101
-----
s(100) = 5050
================
Cheers,
Stan H.
----------------

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Calculation of the following:
1+2+3+4....+97+98+99+100
I tried an arithmetic sequence a=1(100-1)1 but it wont work.

I take it you're trying to sum all integers from 1 to 100, inclusive, with a difference of 1 between terms.
Number of terms (n)in the series: 100
The formula for the sum of a sequence, when the 1st and last terms, the number of terms in the series,
as well as the common difference, are known is: S%5Bn%5D+=+n%28a%5B1%5D+%2B+a%5Bn%5D%29%2F2
Sum of sequence: S%5Bn%5D
Number of terms in series: 100
First term in sequence, or a%5B1%5D: 1
Last term in sequence, or a%5Bn%5D: 100
S%5Bn%5D+=+n%28a%5B1%5D+%2B+a%5Bn%5D%29%2F2 now becomes: S%5B100%5D+=+100%281+%2B+100%29%2F2
S%5B100%5D+=+100%28101%29%2F2
S%5B100%5D+=+50cross%28100%29%28101%29%2Fcross%282%29
The sum of the 1st 100 POSITIVE INTEGERS is: 50(101), or highlight_green%285050%29
You can do the check!!
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