# The third term of an arithmetic sequence is 14, and the ninth term is -1. How do you find the first four terms of the sequence?

Nov 15, 2016

$19 \text{ "16.5" "14" "11.5" } \ldots .$

#### Explanation:

You can write an equation (or formula) for each term in an AP.

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

"The third term is 14" can be written with the formula as:

$\textcolor{b l u e}{{T}_{3} = a + 2 d = 14} \text{ } \rightarrow \left(n - 1\right) = 3 - 1 = 2$

"The ninth term is -1 " can be written as:

$\textcolor{b l u e}{{T}_{9} = a + 8 d = - 1} \text{ } \rightarrow \left(n - 1\right) = 9 - 1 = 8$

Now there are two equations with two unknowns.
Solve them simultaneously (as a system).

color(blue)(T_9= a +8d= -1............................A
$\textcolor{b l u e}{{T}_{3} = a + 2 d = + 14}$ ........................B

A-B: $\text{ } 6 d = - 15$
$\text{ "d = -15/6 = -5/2" } \leftarrow$this is the value for d

$\textcolor{b l u e}{{T}_{3} = a + 2 \left(\frac{- 5}{2}\right) = + 14}$

$a - 5 = 14$
$a = 19 \text{ } \leftarrow$ this is the value for a

This sequence is given by $\textcolor{red}{{T}_{n} = 19 - \frac{5}{2} d}$

${T}_{1} = a = 19$
For ${T}_{2} : n = 2 \rightarrow 19 - \frac{5}{2} \left(1\right) = 16.5$
${T}_{3} : 16.5 - \frac{5}{2} = 14$
${T}_{4} : 14 - \frac{5}{2} = 11.5$

Each term is 2.5 less than the previous term:

$19 \text{ "16.5" "14" "11.5" } 9 \ldots .$