1. An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3, 5, 7, 9, 11 is an arithmetic sequence with five terms.
(a) The first four terms of an arithmetic sequence are $20,13,6,-1$. What are the next two terms in the sequence?
(b) The numbers $2, a, b, c, 14$ in that order form an arithmetic sequence with five terms. What are the values of $a, b$, and $c$ ?
(c) The numbers 7, 15 and $t$, arranged in some unknown order, form an arithmetic sequence with three terms. Determine all possible values of $t$.
(d) The numbers $r, s, w, x, y, z$ in that order form an arithmetic sequence with six terms. The sequence includes the numbers 4 and 20 , with the number 4 appearing before the number 20 in the sequence. Determine the largest possible value of $z-r$ and the smallest possible value of $z-r$.
1. (a) In the arithmetic sequence $20,13,6,-1$, each term is obtained by adding a constant. This constant is $13-20=-7$.
Therefore, the next two terms are $-1+(-7)=-8$ and $-8+(-7)=-15$.
(b) The arithmetic sequence $2, a, b, c, 14$ has 5 terms.
This means that the constant difference (that is, the constant that is added to one term to obtain the next) is added 4 times to the first term (which equals 2) to obtain the last term (which equals 14 ).
Thus, the constant difference must be $\frac{14-2}{4}=\frac{12}{4}=3$.
This means that the sequence starts at 2 and counts by $3 \mathrm{~s}: 2,5,8,11,14$. Therefore, $a=5, b=8$, and $c=11$.
(c) There are six possible ways in which the three terms can be arranged:
$$
7,15, t \quad 15,7, t \quad 7, t, 15 \quad 15, t, 7 \quad t, 7,15 \quad t, 15,7
$$

We determine the corresponding value of $t$ in each case:
- $7,15, t$ : Here, the constant difference is $15-7=8$. Thus, $t=15+8=23$.
- $15,7, t$ : Here, the constant difference is $7-15=-8$. Thus, $t=7+(-8)=-1$.
- $t, 7,15$ : Here, the constant difference is $15-7=8$. Thus, $t+8=7$ and so $t=-1$.
- $t, 15,7$ : Here, the constant difference is $7-15=-8$. Thus, $t+(-8)=15$ and so $t=23$.
- 7, $t, 15$ : Here, the constant difference is added 2 times to 7 to get 15 . Thus, the constant difference is $\frac{15-7}{2}=\frac{8}{2}=4$. Thus, $t=7+4=11$.
- 15, $t, 7$ : Here, the constant difference is $\frac{7-15}{2}=\frac{-8}{2}=-4$. Thus, $t=15+(-4)=11$. Therefore, the possible values of $t$ are $23,-1$, and 11 .
(d) Suppose that the sequence $r, s, w, x, y, z$ has constant difference $d$.
This means that the difference between any two consecutive terms is $d$.
Since the sequence has 6 terms, then the difference is added 5 times to $r$ to get $z$. In other words, $z-r$ is equal to $5 d$.
Also, the difference between a later term in the sequence and an earlier term in the sequence must equal one of $d, 2 d, 3 d, 4 d$, or $5 d$, depending on the number of terms between the two specific terms.
Since 4 and 20 are two terms in the sequence, with 20 appearing later than 4 , and $20-4=$ 16 , then either $d=16$ or $2 d=16$ (which gives $d=8$ ) or $3 d=16$ (which gives $d=\frac{16}{3}$ ) or $4 d=16$ (which gives $d=4$ ) or $5 d=16$ (which gives $d=\frac{16}{5}$ ).
The corresponding values of $5 d$ are $80,40, \frac{80}{3}, 20$, and 16 .
Since $z-r$ is equal to $5 d$, then the largest possible value of $z-r$ is 80 and the smallest possible value is 16 .
(The sequence $4,20,36,52,68,84$ shows that a difference of 80 is possible. The sequence $4, \frac{36}{5}, \frac{52}{5}, \frac{68}{5}, \frac{84}{5}, 20$ shows that a difference of 16 is possible.)