The question is somewhat unclear to me as stated.

Do you mean that you have one arithmetic sequence with difference 10, one with difference 11, and so on?

If that is what you meant, then the answer is no :

Suppose it were possible, then working with the generating functions you have: (*)

x^{a_1} / (1 - x^{10}) + ... + x^{a_{11}} / (1 - x^{20} ) = x^C /(1-x) + extra terms with positive coefficients

( C -> the somewhere you mentioned )

Then multiplying both sides by (1-x) and letting x -> 1^- we get that

1/10 + 1/11 + ... + 1/20 = 1 + something non-negative ... which isabsurdsince the LHS is < 1.

(*) Note here that x^a / (1-x^b) = x^a ( x^{b · 0 } + x^{b · 1} + ...) = x^{a + b · 0} + x^{a + b · 1} ...

which corresponds to the arithmetic sequence starting at 'a' and with difference 'b'.

Sorry for the disgusting looking post, but latex appears not to be working properly.