# How do you find the 9th term of the arithmetic sequence a10 = 100 and a20 = 50?

Mar 15, 2018

${a}_{9} = 105$

#### Explanation:

Given: arithmetic sequence: a_10 = 100; a_20 = 50

An arithmetic sequence has the form: ${a}_{n} = {a}_{1} + d \left(n - 1\right)$, where ${a}_{1}$ is the first number in the sequence, $n$ is the number in the sequence and $d$ is the arithmetic difference between consecutive terms: $d = {a}_{2} - {a}_{1} = {a}_{3} - {a}_{2} \ldots$.

${a}_{10} = {a}_{1} + d \cdot \left(10 - 1\right) = 100$
${a}_{10} = {a}_{1} + 9 d = 100$

${a}_{20} = {a}_{1} + d \cdot \left(20 - 1\right) = 50$
${a}_{20} = {a}_{1} + 19 d = 50$

Find $d$:
We have two equations and two unknowns. Use elimination.
Multiply the second equation by $- 1$ and add the two equations:

Equation 1: $\text{ } {a}_{1} + 9 d = 100$
Equation 2 *(-1) = $\underline{+ - {a}_{1} - 19 d = - 50}$
$\text{ } - 10 d = 50$

$d = - 5$

Find ${a}_{1}$ by substitution: $\text{ } {a}_{1} + 19 \cdot - 5 = 50$

Simplify: $\text{ } {a}_{1} - 95 = 50$

${a}_{1} = 50 + 95 = 145$

CHECK: values of ${a}_{1}$ and $d$
a_10 = 145 + 9*(-5) = 100

a_20 = 145 + 19*(-5) = 50

Find ${a}_{9}$:
${a}_{9} = {a}_{1} + 8 d = 145 + 8 \left(- 5\right) = 105$

Yes, because ${a}_{9}$ is slightly larger than ${a}_{10}$ which is larger than ${a}_{20}$