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For any EVENLY SPACED set:[email protected] wrote:The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?
(A) 12
(B) 17
(C) 25
(D) 33
(E) 50
sum = (number of values)(median).
Let x = the median of the four consecutive ODD integers.
Sum = number*median = 4x.
Let y = the median of the three consecutive EVEN integers.
Sum = number*median = 3y.
Since the two sums are equal, we get:
4x = 3y.
y = (4/3)x.
Since y must be an integer, x must be a multiple of 3.
Since the median of 4 consecutive odd integers must be even -- for example, the median of {1, 3, 5, 7} is 4 -- x must be even.
Implication:
x must be an even multiple of 3.
In other words:
x must be a MULTIPLE OF 6.
Let x = 6a, where a is an integer.
Then y = (4/3)(6a) = 8a, where a is an integer.
Implication:
y must be a MULTIPLE OF 8.
Since y must be between 101 and 200, list the multiples of 8 between 101 and 200:
104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192.
Total options = 12.
The correct answer is A.
