Question 854950
If {{{n}}}= the third number, the numbers are
{{{n-4}}} , {{{n-2}}} , {{{n}}} , {{{n+2}}} , {{{n+4}}} , and {{{n+6}}} .
Their sum is
{{{(n-4)+(n-2)+n+(n+2)+(n+4)+(n+6)=6n-4-2+2+4+6)=6n+6}}}
{{{6n+6=366}}}
{{{6n=366-6}}}
{{{6n=360}}}
{{{n=360/6}}}
{{{highlight(n=60)}}}
 
NOTES:
1) The teacher may expect some talk about arithmetic sequences, but it is easy enough as done above.
2) As mental math (as I would do for a multiple choice test),
I know that the sum of consecutive terms in an arithmetic sequence is the average times the number of terms. That average would be {{{366/6=61}}} . I also know that in an arithmetic sequence the average is the median, and with 6 terms that median would be halfway between the third and fourth terms, so the third and fourth terms should be 60 and 62.
3) If I read "consecutive even numbers" or "consecutive odd numbers", I simply call some convenient number {{{n}}} (not caring if {{{n}}} is odd or even until the end). I just make the consecutive even or odd numbers 2 units apart from one another. At the end, the results should be even or odd as required, or else the problem has no answer (or I made a mistake).