Question 1197953
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Option A has the sequence: 1000, 999, 998, ..., 3, 2, 1


It is arithmetic with
{{{a[1]}}} = 1000 = first term
d = -1 = common difference
n = 1000 terms


{{{S[n]}}} = sum of the first n terms of an arithmetic sequence


{{{S[n] = (n/2)*(2*a[1]+d(n-1))}}}


{{{S[1000] = (1000/2)*(2*1000-1(1000-1))}}}


{{{S[1000] = 500500}}}
Therefore, adding the terms 1000,999,998,...,3,2,1 will get us the sum 500500
1000+999+998+...+3+2+1 = 500500


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Option B is a geometric sequence: 1000, 1000*(1.12), 1000*(1.12)^2, ..., 1000*(1.12)^(36-1)


Note: 3 years = 3*12 = 36 months


a = 1000 = first term
r = 1.12 = common ratio, to represent an increase of 12%
n = 36 = number of terms


{{{S[n]}}} = sum of the first n terms of a geometric sequence


{{{S[n] = (a*(1-r^n))/(1-r)}}}


{{{S[36] = (1000*(1-(1.12)^36))/(1-1.12)}}}


{{{S[36] = 484463.11607167}}}


{{{S[36] = 484463.12}}}


If you go for option A, then you'll get a total of $500,500.
If you go for option B, then you'll get a total of $484,463.12


We see that option A is better by 16036.88 dollars since 500500-484463.12 = 16036.88


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Answer: Option A offers the most money at $500,500.
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