Sequence

[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

For any EVENLY SPACED set:
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.

[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

The sum of 3 consecutive EVEN integers must be equal to the sum of 4 consecutive ODD integers.

Alternate approach:
TEST CASES with easier values and LOOK FOR A PATTERN.

2, 4, 6, 8, 10, 12...
1, 3, 5, 7, 9, 11...
The red sum is equal to the blue sum:
6+8+10 = 3+5+7+9.

...8, 10, 12, 14, 16, 18, 20...
...7, 9, 11, 13, 15, 17...
The red sum is equal to the blue sum:
14+16+18 = 9+11+13+15.

Notice the pattern:
The red medians -- 8 and 16 -- are CONSECUTIVE MULTIPLES OF 8.
The blue medians -- 6 and 12 -- are CONSECUTIVE MULTIPLES OF 6.
Test whether the next greatest multiples of 8 and 6 -- 24 and 18 -- are viable options for the two medians:
22, 24, 36
15, 17, 19, 21
Success!
The two sums are equal:
22+24+26 = 15+17+19+21.

Implication:
For the sum of 3 consecutive even integers to be equal to the sum of 4 consecutive odd integers, the median of the 3 even integers must be a multiple of 8, while the median of the 4 odd integers must be a multiple of 6.
Since the median of the 3 even integers 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.