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July 31, 2014

July 31, 2014

Posted by **anonymous** on Saturday, March 8, 2008 at 1:08pm.

- math -
**Count Iblis**, Saturday, March 8, 2008 at 1:20pmYou can make use of:

Sum from n = 0 to N of n = 1/2 N (N+1)

- math -
**tchrwill**, Saturday, March 8, 2008 at 4:26pmn.....1.....2.....3.....4.....5.....6

N....-10...-6....-2.....2.....6.....10

The 120th term derives from L = a + (n - 1)d where a = the first term, n = the number of terms and d = the common difference.

Then, the sum is S = n(a + L)/2.

- math -
**tchrwill**, Saturday, March 8, 2008 at 4:26pmn.....1.....2.....3.....4.....5.....6

N....-10...-6....-2.....2.....6.....10

The 120th term derives from L = a + (n - 1)d where a = the first term, n = the number of terms and d = the common difference.

Then, the sum is S = n(a + L)/2.

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