the first three terms of an arithmetic sequence are x-2;2x+6 and 4x-8 respectively..determine.....(1)x (2) the 20th term (3) the sum of the first 20 terms of the sequence

Answer 1

1)T2-T1=T3-T2
2x+6-(x-2)=4x-8-(2x+6)
2x+6-x+2=4x-8-2x-6
x+8=2x-14
x=22
2)substitute for x:
1st term:x-2=22-2=20
2nd term:2x+6=2(22)+6=44+6=50
3rd term:4x-8=4(22)-8=88-8=80
now we look for the constant difference(d):
T2-T1=50-20=30
T3-T2=80-50=30
So d=30
Tn=an+(n-1)d
T20=20(20)+(20-1)(30)
T20=400+(19)(30)
T20=400+570
T20=970
3)Sn=n/2(2a+(n-1)d)
S20=20/2[2(20)+(19)(30)]
s20=10(40+570)
s20=10(610)
s20=6100