# What would the second and third terms of an arithmetic sequence be if the first term is 6 and the fourth is 33?

May 16, 2018

The second term = 15, and the third term = 24

#### Explanation:

An arithmetic sequence can be written on the form
${s}_{n} = a n + b$
We know that ${s}_{1} = 6$ and ${s}_{4} = 33$
Therefore: ${s}_{4} - {s}_{1}$=(4a+b)-(a+b)$= 33 - 6 = 27$
$3 a = 27$
$a = 9$
As $6 = 9 + b$,
$b = - 3$
Therefore ${s}_{n} = 9 n - 3$
We, therefore, get:
${s}_{1} = 1 \cdot 9 - 3 = 6$
${s}_{2} = 2 \cdot 9 - 3 = 15$
${s}_{3} = 3 \cdot 9 - 3 = 24$
${s}_{4} = 4 \cdot 9 - 3 = 33$