Question 216836
Can you please help me find four consecutive odd integers that have a sum of 440.


Step 1.  Let n be one odd integer.


Step 2.  Let n+2, n+4, and n+6 be the next three odd and consecutive integers.


Step 3.  Then n+n+2+n+4+n+6=440 since four consecutive odd integers that have a sum of 440.


Step 4.  Simplifying and olving the equation in Step 3 yields the following steps:


{{{n+n+2+n+4+n+6=440}}}


{{{4n+12=440}}}


Subtract 12 from both sides of the equation yields


{{{4n+12-12=440-12}}}


{{{4n=428}}}


Divide 4 to both sides of the equation


{{{4n/4=428/4}}}


{{{n=107}}} 


Then  {{{n+2=109}}} {{{n+4=111}}} {{{n+6=113}}} 


Check sum...107+109+111+113=440...which is a true statement.


Step 5.  The 4 consecutive odd integers are:  107, 109, 111, and 113.


I hope the above steps were helpful. 


For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.


And good luck in your studies!


Respectfully,
Dr J