# How do you write a rule for the nth term of the arithmetic sequence given a_6=13, a_14=25?

Apr 19, 2018

$\textcolor{b l u e}{4 + \frac{3}{2} n}$

#### Explanation:

The nth term of an arithmetic sequence is given as:

$a + \left(n - 1\right) d$

Where $\boldsymbol{a}$ is the first term, $\boldsymbol{d}$ is the common difference and $\boldsymbol{n}$ is the nth term.

We need to find the first term and the common difference:

${a}_{6} = 13 \implies a + \left(6 - 1\right) d = 13$

${a}_{14} = 25 \implies a + \left(14 - 1\right) d = 25$

$a + 5 d = 13 \setminus \setminus \setminus \setminus \left[1\right]$

$a + 13 d = 25 \setminus \setminus \setminus \setminus \setminus \left[2\right]$

Solving simultaneously:

Subtract $\left[1\right]$ form $\left[2\right]$

$0 + 8 d = 12 \implies d = \frac{3}{2}$

Substituting this in $\left[1\right]$

$a + 5 \left(\frac{3}{2}\right) = 13$

$a = 13 - 3 \left(\frac{3}{2}\right) = \frac{11}{2}$

For nth term:

$\frac{11}{2} + \left(n - 1\right) \left(\frac{3}{2}\right) = \frac{11}{2} + \frac{3}{2} n - \frac{3}{2} = \textcolor{b l u e}{4 + \frac{3}{2} n}$