E G F A G B A Every other letter is in alphabetical
g.
g
If you read Ludwig Wittgenstein's comments on language games, you will find that any next-in-series problem has an infinite number of soluitons. I am curious, though, to learn what you or others think is the "uniquely correct" answer to your challenge.
(F+G)/2
The next character in the series, "AB 3 D 56 G" is 8.
G C C D D E C E G G F E D F D D E E F D E G F E D C E G E G E G E E G G F E D D F F D F D D G F E D C And it is the same for the next two veruses
a-g-f-f-f-f-f-f-f-g-a-g-a-g-f-f-f-f-f-f-f-f-f-a-g-g-a-g-f-f-g-a-a-a-a-a-a-g-f
Here is the pattern. It begins with the first letter of the alphabet and then "adds" five letters. The next letter comes from the end of the alpahabet and "subtracts" five letters. The fifth in the series (G) adds one off from the second (F). If one adds five to that the sixth in the series (L) comes next. We then move to the letter before "U" to get the seventh in the series. To sum it all: We added five to get from the A to F and G to L. We subtracted five to get from Z to U. To answer the question, we subtract five from T to get O I think!
T
G G G G F# G G F# G A B A G G G G F# G G low D (ALL THIS REPEATS) ( THE NEXT PART DON'T REPEAT) G A LOW D HIGH D C B A B C B A G F# G D E G A LOW D HIGH D C B A B C B A G F# G F# G A B A G
a-g-f-f-f-f-f-f-f-g-a-g-a-g-f-f-f-f-f-f-f-f-f-a-g-g-a-g-f-f-g-a-a-a-a-a-a-g-f that is only chorus