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April 20, 2014

April 20, 2014

Posted by **t.t** on Tuesday, March 27, 2012 at 5:21pm.

- Maaaaaaaath -
**Piwo**, Tuesday, April 10, 2012 at 1:00pm1)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

- Maaaaaaaath -
**Piwo**, Tuesday, April 10, 2012 at 1:00pm1)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

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