# What is the simple formula for the nth term for 10, 14, 18, 22, 24, ...?

Jul 15, 2015

${a}_{n} = 6 + 4 n$ for $n = 1 , 2 , 3 , 4 , \ldots$

#### Explanation:

Probably the simplest approach is to look for a pattern. If you let ${a}_{1} = 10$ be the first term, ${a}_{2} = 14$ be the second term, ${a}_{3} = 18$ be the third term, ${a}_{4} = 22$ be the fourth term, etc..., you should note that ${a}_{2} = {a}_{1} + 4 = 10 + 4$, ${a}_{3} = {a}_{2} + 4 = {a}_{1} + 8 = {a}_{1} + 2 \cdot 4 = 10 + 2 \cdot 4$, ${a}_{4} = {a}_{3} + 4 = {a}_{1} + 2 \cdot 4 + 4 = {a}_{1} + 3 \cdot 4 = 10 + 3 \cdot 4$, etc...

From this, I hope you see that the pattern is ${a}_{n} = 10 + \left(n - 1\right) \cdot 4$. This is the formula for the ${n}^{t h}$ term. It can be simplified to ${a}_{n} = 6 + 4 n$ (for $n = 1 , 2 , 3 , 4 , \ldots$)

You can check that this works. For instance, ${a}_{5} = 6 + 4 \cdot 5 = 26$ (you made a typo when you wrote 24) and ${a}_{6} = 6 + 4 \cdot 6 = 30$. You can even find the millionth term, or any other, this way: ${a}_{1000000} = 4000006$.

This is an example of an arithmetic sequence . In general, an arithmetic sequence takes the form ${a}_{1} , {a}_{2} = {a}_{1} + d , {a}_{3} = {a}_{2} + d = {a}_{1} + 2 d , {a}_{4} = {a}_{3} + d = {a}_{1} + 3 d , \ldots$ for some numbers ${a}_{1}$ (the first term) and $d$ (the common difference). The pattern shows that ${a}_{n} = {a}_{1} + \left(n - 1\right) d = \left({a}_{1} - d\right) + \mathrm{dn}$ for $n = 1 , 2 , 3 , \ldots$ This is a (discrete) linear function of $n$.

For extra fun, we can find the sum of the first $N$ terms of such an arithmetic sequence can be found by using the fact that $1 + 2 + 3 + \setminus \cdots + \left(N - 1\right) = \frac{\left(N - 1\right) N}{2}$ (check this!):

${a}_{1} + {a}_{2} + {a}_{3} + \setminus \cdots + {a}_{N}$

$= {a}_{1} + \left({a}_{1} + d\right) + \left({a}_{1} + 2 d\right) + \setminus \cdots + \left({a}_{1} + \left(N - 1\right) d\right)$

$= \left({a}_{1} + {a}_{1} + {a}_{1} + \setminus \cdots + {a}_{1}\right) + \left(d + 2 d + 3 d + \setminus \cdots + \left(N - 1\right) d\right)$

$= {a}_{1} N + d \cdot \frac{\left(N - 1\right) N}{2} = \frac{d}{2} {N}^{2} + \left({a}_{1} - \frac{d}{2}\right) N$

$= \frac{N}{2} \left(2 {a}_{1} + d \left(N - 1\right)\right)$

Since ${a}_{N} = {a}_{1} + d \left(N - 1\right)$, this can also be written as $\frac{N}{2} \left({a}_{1} + {a}_{N}\right) = N \cdot \frac{{a}_{1} + {a}_{N}}{2}$ or N*(\mbox{average of the first and last terms)