Is the arithmetic sequence a subspace?

Question

Let a and k be some constants. Is $(a, a+k, a+2k...)$ a subspace of the set of all infinite sequences of real numbers?

My book says it is, but I don't understand why. Let $k>2*a$. Hence, 2a is not within the set, so this set shouldn't be a subspace.

Answer 1

Proof sketch/idea.

Let $S$ be the set of all sequences of real numbers. It's a vector space with the obvious (I hope) operations: add corresponding entries, multiply all entries by a scalar.

Let $A$ be the subset of arithmetic progressions.

Is the zero vector in $S$ a member of $A$?

If you add two sequences in $A$ do you get another element of $A$?

If you multiply a sequence in $A$ by a scalar is the result in $A$?