guerrero wrote:For every positive integer n, the nth term of a sequence is the total of three consecutive integers starting at n. What is the total of terms 1 through 99 of this series?
A. 5,250
B. 10,098
C. 14,850
D. 15,147
E. 15,150
OA D
WRITE IT OUT until you see the PATTERN.
First 4 terms of the sequence:
1+2+3
2+3+4
3+4+5
4+5+6
Last 4 terms of the sequence:
96+97+98
97+98+99
98+99+100
99+100+101
The first value (1) and the last value (101) each appear only once.
1+101 = 102.
The second value (2) and the penultimate value (100) each appear two times.
2+2+100+100 = 204.
Every integer between 3 and 99, inclusive, appears THREE TIMES.
For any set of consecutive integers:
The number of integers = biggest - smallest + 1.
The average of the integers = (biggest + smallest)/2.
The sum of the integers = number * average.
Thus, for the consecutive integers between 3 and 99, inclusive:
Number = 99-3+1 = 97.
Average = (99+3)/2 = 51.
Sum = 97*51 = 4947.
Since each of these integers appears three times, we multiply by 3:
3 * 4847 = 14,841.
Thus:
Total sum = 102 + 204 + 14841 = 15,147.
The correct answer is
D.
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