2007-12-12 01:57:43 · 8 answers · asked by john a 2 in Science & Mathematics ➔ Mathematics
Sn = (n/2) [ 2a + (n-1) d ]
n = 10
d = - 1
a = 10
Sn = 5 [ 20 - 9 ]
Sn = 55
2007-12-12 02:44:00 · answer #1 · answered by Como 7 · 4⤊ 0⤋
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RE:
What is the formula for adding a group of descending sequential numbers i.e. 10+9+8+7+6+5+4+3+2+1?
2015-08-18 12:44:34 · answer #2 · answered by ? 1 · 0⤊ 0⤋
Adding Sequential Numbers
2016-12-28 10:02:45 · answer #3 · answered by harabedian 4 · 0⤊ 0⤋
Let the sequence be n+(n-1)+(n-2) ........ +2+1
The sum is given by the formula
( n(n+1) )/2
where n is the beginning number.
Eg. Take the case with the sequence starting with 10.
The sum would be (10 * (10+1) ) /2 = 55
2007-12-12 02:09:42 · answer #4 · answered by w4c~m3-5un 3 · 2⤊ 0⤋
Descending Numbers
2016-10-20 05:18:40 · answer #5 · answered by inmon 4 · 0⤊ 0⤋
The formula for adding the first n digits is (n)(n+1)/2.
In this case (10)(11)/2 = 55.
2007-12-12 02:03:55 · answer #6 · answered by stanschim 7 · 3⤊ 1⤋
What does it matter if they are descending or ascending!
Here is the formula to add the first n positive integers i.e.
1+2+...+n = n(n+1)/2, so in your case, n=10 and the answer is 55.
2007-12-12 02:06:26 · answer #7 · answered by fouman1 3 · 2⤊ 1⤋
(u(u+1)) / 2 where u is your starting number
2016-07-02 13:49:53 · answer #8 · answered by maths 1 · 0⤊ 0⤋
anyone please help me find an calculator for this formula, I try find all day and only find for ascending numbers :(
2015-12-15 10:11:57 · answer #9 · answered by Larisa 1 · 0⤊ 0⤋