Find the sum of an arithmetic series in which a1=12, a4=-6, and n=10.
2007-12-17 18:33:38 · 3 answers · asked by rbrulesamanda 2 in Science & Mathematics ➔ Mathematics
Greetings
a1 = 12
an = 12 +(n -1)d where an is the nth term and d is the constant difference
a4 = 12 + 3d = -6
3d = -18
d = -6
a10 = 12 +9d = 12 - 6*9 = -42
sum = (first + last)*#ofterms/2 = (-30)*10/2 = -150
Regards
2007-12-17 18:45:58 · answer #1 · answered by ubiquitous_phi 7 · 0⤊ 0⤋
n th term = a + (n - 1) d
a = 12
a + 3d = - 6
12 + 3d = - 6
3d = - 18
d = - 6
Sn = (n/2) [ 2a + (n - 1) d ]
S10 = 10/2 [ 24 + (9)(- 6) ]
Sn = 5 [- 30 ]
Sn = - 150
2007-12-17 22:20:11 · answer #2 · answered by Como 7 · 3⤊ 0⤋
theres a formula for this,
i think it was something like
Sn = 2a1 + (n-1)d
2007-12-17 18:41:49 · answer #3 · answered by G3P 3 · 0⤊ 0⤋