Answer: A
A:
Sn = Sn-1 + 3/2 and S1 = 2
Sn = a + (n-1)d
a = S1 (the first element in the sequence)
d = Sn - Sn-1 = 3/2 (distance between two following elements in this kind of sequence which every new element is made by adding a constant value to it’s previous element)
Sn = 2 + (n-1) * 3/2 = 2 + 3/2*n - 3/2 = 1/2 + 3/2*n
Sum of the terms in S from s1 to S13, inclusive = (S1+ S13) *n / 2 = (2+20)*13/2=11*13 = 143
S1 = 2
S13 = 1/2 + 3/2*13 = 20
B:
The sum of the terms in A from A1 to A13, inclusive =
An = An-1 -1/2 and S1 = 18.5
Sn = a + (n-1)d
a = A1 (the first element in the sequence)
d = Sn - Sn-1 = -1.5 (distance between two following elements in this kind of sequence which every new element is made by adding a constant value to it’s previous element)
An = 18.5 + (n-1) * (-1.5) = 18.5 - 3/2*n + 3/2 = 20- 3/2*n
Sum of the terms in A from A1 to A13, inclusive = (A1+A13) *n / 2 = (18.5 + 0.5)*13/2=19*13/2 = 123.5
A1 = 18.5
A13 = 20-3/2*13 = 1/2
So A is bigger.
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