How do you find the first five terms of an airthmetic series with
t12=35 and S20=610 ( the 12 and 20 are superscriped)
Please explain thanks :)
2007-06-20 19:23:03 · 3 answers · asked by fast.lane17 1 in Science & Mathematics ➔ Mathematics
let the first term be a and the common difference be d
then t12=a+11d=35
s20=20/2(2a+(20-1)d)=610
so we get 2a+19d=61
from here get d=3 and a=2
so the first five terms are 2,5,8,11,14
2007-06-20 19:33:58 · answer #1 · answered by Anonymous · 1⤊ 0⤋
The 12th term is 35 and the sum of the frst 20 terms is 610
Let the first term be a and the common difference be d
a + 11d = 35 ..... (1)
10(2a + 19d) = 610
2a + 19d = 61.....(2)
(1)*2 = 2a + 22d = 70 .... (3)
Subtract (2) from (3),
3d = 9
d = 3
Put d = 3 in (1)
a + 33 = 35
a = 2
The first 5 terms are a, a + d, a + 2d, a + 3d, a + 4d
2, 5, 8, 11, 14
2007-06-20 19:45:49 · answer #2 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
t12 = a + 11d
a + 11d = 35 (i)
S20 = 10 (2a + 19d)
20a + 190 d = 610 (ii)
Multiply (i) by 20 and subtract (ii) from it :
220d - 190d = 30d = 90
d = 3
Put in (i) -
a = 2
T1 = a = 2
T2 = a + d = 5
T3 = 8
T4 = 11
T5 = 14
Hope this helps.
your_guide123@yahoo.com
2007-06-20 19:30:14 · answer #3 · answered by Prashant 6 · 2⤊ 0⤋