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?  Select.
A) 12	B) 17	C) 25	D) 33	E) 50
Sum of four consecutive odd numbers:   (2a - 3) + (2a - 1) + (2a + 1) + (2a + 3) = 8a   Sum of three consecutive even numbers:   (2b - 2) + 2b + (2b + 2) (2  - 3) + (2  - 1) + (2  + 1) + (2  + 3)= 
<gadget id="calculator">(2  - 3) + (2  - 1) + (2  + 1) + (2  + 3)</gadget>
<output>8</output>
8a   Sum of three consecutive even numbers:   (2b - 2) + 2b + (2b + 2) = 6b   Given 8a (2  - 2) + 2  + (2  + 2)= 
<gadget id="calculator">(2  - 2) + 2  + (2  + 2)</gadget>
<output>6</output>
6b   Given 8a= 6b or a/b = 3/4, a and b can be any integers. So, 'a' has to be a multiple of 3 and 'b' has to be a multiple of 4. Possible solutions are: a = 3, b = 4; a = 6, b = 8; a = 9, b = 12 etc   Since 101 &lt; 2b &lt; 200 i.e.   51 &lt;= b &lt; 100   Since b also has to be a multiple of 4, the values that b can take are 52, 56, 60, 64 ... 96   Number of values b can take = (Last term - First term)/Common Difference + 1 = (96 - 52)/4 + 12   12nswer:12
<result>A</result>
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?