# Given 10, 17, 24, 31, the nth term is given by 7n+3. How you know?

Dec 4, 2016

By realizing that $10 = 7 \left(1\right) + 3$, $17 = 7 \left(2\right) + 3$, etc.

An arithmetic sequence is one where some constant difference between numbers in a sequence gives you a pattern to get to the next unlisted number in the sequence.

In this case, the difference is $7$. When one works backwards and subtracts $7$ from $10$, the result is $3$. At the $0$th term, we assume that $n > 0$ for series and sequences.

So, $3$ is the $0$th term, giving us the y-intercept, so to speak. Now we would simply have to recognize that if we increment $n$ integrally, and we have a difference of $7$ from term to term, that is the operation $7 n$.

So, the arithmetic sequence would be generated as:

$\textcolor{g r e e n}{{a}_{0}} = 3 + 7 \left(0\right) = \textcolor{g r e e n}{3}$

$\textcolor{g r e e n}{{a}_{1}} = {a}_{0} + 7 \left(1\right) = \textcolor{g r e e n}{10}$

$\textcolor{g r e e n}{{a}_{2}} = {a}_{1} + 7 \left(1\right) = {a}_{0} + 7 \left(1\right) + 7 \left(1\right) = \textcolor{g r e e n}{17}$

$\textcolor{g r e e n}{{a}_{3}} = {a}_{2} + 7 \left(1\right) = {a}_{1} + 7 \left(1\right) + 7 \left(1\right)$

$= {a}_{0} + 7 \left(1\right) + 7 \left(1\right) + 7 \left(1\right) = \textcolor{g r e e n}{24}$

and so on. Each term has ${a}_{0}$ within it, added to $7$ times the current index indicated by ${a}_{n}$. So, we just have the recursive definition:

$\textcolor{b l u e}{{a}_{n} = {a}_{n - 1} + 7}$

or the overall definition:

$\textcolor{b l u e}{{a}_{n} = 3 + 7 n}$