Maaaaaaaath
posted by t.t on .
the first three terms of an arithmetic sequence are x2;2x+6 and 4x8 respectively..determine.....(1)x (2) the 20th term (3) the sum of the first 20 terms of the sequence

1)T2T1=T3T2
2x+6(x2)=4x8(2x+6)
2x+6x+2=4x82x6
x+8=2x14
x=22
2)substitute for x:
1st term:x2=222=20
2nd term:2x+6=2(22)+6=44+6=50
3rd term:4x8=4(22)8=888=80
now we look for the constant difference(d):
T2T1=5020=30
T3T2=8050=30
So d=30
Tn=an+(n1)d
T20=20(20)+(201)(30)
T20=400+(19)(30)
T20=400+570
T20=970
3)Sn=n/2(2a+(n1)d)
S20=20/2[2(20)+(19)(30)]
s20=10(40+570)
s20=10(610)
s20=6100