Question 1204430
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a = first installment = first term of arithmetic sequence


n = 10 = number of payments
d = -8 = common difference


Sn = sum of first n terms of arithmetic sequence, aka arithmetic progression (AP)
Sn = 1000


Sn = (n/2)*(2*a + d(n-1))
1000 = (10/2)*(2*a - 8(10-1))
1000 = 5(2a-72)
2a-72 = 1000/5
2a-72 = 200
2a = 200+72
2a = 272
a = 272/2
a = 136


The arithmetic sequence of 10 payments is
136, 128, 120, 112, 104, 96, 88, 80, 72, 64
I used the "sequence" command in GeoGebra to generate this list quickly. A spreadsheet is another good option.


As a check,
136+128+120+112+104+96+88+80+72+64 = 1000
We have confirmed the correct first payment.
An interesting thing to notice is:  
(n/2)*(1st term + 10th term) = (10/2)*(136+64) = 1000



Answer: <font color=red size=4>136</font>
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