it is for algebra 2 and pre-calculus.
2007-05-17 15:28:41 · 5 answers · asked by karakar989 1 in Science & Mathematics ➔ Mathematics
S(n) = n/2 (t1 + tn),
where S(n) is the sum of the first n terms, t1 is the first term and tn is the nth term
2007-05-17 15:32:14 · answer #1 · answered by Kathleen K 7 · 0⤊ 0⤋
It is also
S(n) = (n/2) * [2*t1 + (n-1)*d]
where d = difference between terms
2007-05-17 15:37:43 · answer #2 · answered by Dr D 7 · 0⤊ 0⤋
Let "a" be first term and "d" be common difference.
Sn is sum to n terms:-
-------1st----2nd------3rd----------nth
Sn = a + (a + d) + (a + 2d) ---(a + (n - 1).d)
Sn = (a + (n - 1).d)-------(a + 2d) + (a + d) + a
Adding gives:-
2 Sn = n . [ 2a + (a + (n - 1).d) ]
Sn = (n/2).[ 2a + (a + (n - 1).d ]
2007-05-18 02:50:38 · answer #3 · answered by Como 7 · 0⤊ 0⤋
an = a1 + (n - 1)d
first find out what "an" gets you, then you can plug in everything that is needed.
Sn = (n/2)(a1 + an)
2007-05-17 16:26:47 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋
a+2a+3a+4a+.......
nth term =a+(n-1)d
sum to n terms: S(n) = (n/2)*[2a+ (n-1)d]
2007-05-17 15:37:41 · answer #5 · answered by Pavan 3 · 0⤊ 0⤋