arithmetic sequence - find if a number is in a sequence Given the following 2 arithmetic sequences:


*

*Sequence: 4, 7, 10, 13, ...

*Sequence: 3, 6, 9, 12, ...


I know that 3n + 1 = 22 will be in sequence 1 and not in sequence 2. Both have a common difference of 3, but different starting points.
What is the equation to prove that 22 is part of sequence 1 and not 2 though? Right now, all 3n + 1 = 22 tells me is whether a number exists where the common difference is 3. Sequence 2, however, has a common difference of 3. 3n + 1 == 22, doesn't tell me that 22 is not in sequence 2. I need an equation to prove, where the starting point of a sequence is 3 or 4, etc., with a common difference of 3, that a number is actually in the given sequence.
 A: Your statement that $3n+1=22$ is in this sequence:
$ 4, 7, 10, 13, 16, 19, 22,...$ is not correct. The reason is that $n$ is usually used to denote the position of the element in the sequence.
The following is true for any term after the first in an arithmetic sequence where a=first term and r is difference between any 2 consecutive terms $(n>1)$, i.e. the sequence looks like: $$a, a+r, a+2r,...$$ etc.
General Term = N-th Term: $$a_{n}=a+r(n-1)$$
In the first sequence you have a=4 and r=3, so:
$$a_2=4+3(2-1)=7$$
$$a_3=4+3(3-1)=10$$
$$a_4=4+3(4-1)=13$$
...

What is the equation to prove that 22 is part of sequence 1

We have to use the general term formula again and test an valid integer (n) is a solution for:
$$22=4+3(n-1)$$
This is:
$$22=4+3(n)-3$$
Which leads to $n=7$ which is the correct value. As a result we can tell that 22 belongs to this sequence.
For the second sequence, you have a 
$$3, 6, 9, 12, ...$$
This tells us that $a=3$ and $r=3$, so 
$$a_{n}=3+3(n-1)$$
You could set:
$$22=3+3(n-1)$$
and solve for n to get $n=\frac{22}{3}$ which is not a position in the sequence at least since it is not an integer. This proves that $22$ is not part of the 2nd series.

I need an equation to prove, where the starting point of a sequence is
  3 or 4, etc., with a common difference of 3, that a number is actually
  in the given sequence.

given a vlaue $k$, to prove that $k$ is part of a given sequence, solve the following equation for $n$ and check to see that $n$ is an integer >=1: $$n=\frac{(k+r)-a}{r} $$
A: Working modulo $3$, both sequences are constants. Therefore any number greater than or equal to the first term in the sequence that is equivalent to every term modulo $3$ is within the sequence. For example, the first sequence is just $1 \text{ mod }3$ hence the given number $22\ge4$ is in the sequence as $22\equiv1\text{ mod }3$. The second sequence is $0\text{ mod }3$ so $22$ cannot be within the second sequence.
A: Sequence 2 is given by $3n$, where $n = 1, 2, 3,..$. Since 22 is not divisible by 3, it shouldn't be an element of sequence 2.
