arithmetic sequence - find if a number is in a sequence

Question

Given the following 2 arithmetic sequences:

1. Sequence: 4, 7, 10, 13, ...
2. 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.

Answer 1

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: $$n=\frac{(k+r)-a}{r} $$

Answer 2

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.

Answer 3

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.