# How do you write a rule for the nth term of the arithmetic sequence given a_9=-14, a_15=-20?

Feb 8, 2018

#### Answer:

${a}_{n} = - \left(6 + \left(n - 1\right)\right)$

#### Explanation:

General equation of the ${n}^{t h}$ term of an arithmetic sequence is

${a}_{n} = {a}_{1} + \left(n - 1\right) d$

${a}_{9} = {a}_{1} + \left(9 - 1\right) d = - 14$ Eqn (1)

${a}_{15} = {a}_{1} + \left(15 - 1\right) d = - 20$ Eqn (2)

Subtraction Eqn(1) from (2),

$14 d - 8 d = - 6$

$6 d = - 6$ or $d = - 1$

Substituting the value of d in Eqn (1),

${a}_{1} + \left(9 \cdot - 1\right) = - 14$

${a}_{1} = - 14 + 8 = - 6$

${a}_{n} = - 6 + \left(\left(n - 1\right) \cdot - 1\right) = - \left(6 + \left(n - 1\right)\right)$