Answer:
A. -200
Step-by-step explanation:
The sum of n terms of an arithmetic sequence is given by the formula ...
Sn = (2·a1 +d(n -1))(n/2)
where a1 is the first term of the sequence, and d is the common difference.
This problem requires we find the sum of 4 terms of a sequence that begins 100, 88, 76, .... Those terms are the 12th through the 15th.
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The general term of an arithmetic sequence is given by the formula ...
an = a1 +d(n-1)
where a1 is the first term and d is the common difference.
The given sequence has first term a1=100, and common difference ...
d = 88 -100 = -12
Then the general term is ...
an = 100 -12(n -1)
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We want the sum of terms 12–15. The first term of this sub-sequence is ...
a12 = 100 -12(12 -1) = 100 -132 = -32
The next terms of the sequence will have the same common difference, -12. We can use the given sum formula to find the sum of the four terms:
Sn = (2·a1 +d(n -1))(n/2) . . . . . . . for a1=-32, d=-12, and n=4
S4 = (2·(-32) -12(4 -1))(4/2) = (-64 -36)(2) = -200
The sum of the 12th to 15th terms of the given sequence is -200.
