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, $5,7,9$ is an arithmetic sequence with three terms.

(a) Three of the five numbers $2,5,10,13,15$ can be chosen to form an arithmetic sequence with three terms. What are the three numbers?

(b) The numbers $p, 7, q, 13$, in that order, form an arithmetic sequence with four terms. Determine the value of $p$ and the value of $q$.

(c) The numbers $a, b, c,(a+21)$, in that order, form an arithmetic sequence with four terms. Determine the value of $c-a$.

(d) The numbers $(y-6),(2 y+3),\left(y^{2}+2\right)$, in that order, form an arithmetic sequence with three terms. Determine all possible values of $y$.
(a) From the given list, the numbers 5, 10, 15 form an arithmetic sequence with common difference 5 . None of the other sets of three numbers form an arithmetic sequence.

(b) Solution 1

Moving from the 2nd term to the 4th term involves adding the common difference twice. Since the difference between the 2 nd term and the 4 th term is $13-7=6$, then 2 times the common difference equals 6 , which means that the common difference is 3 .

Therefore, the 3rd term, $q$, is 3 more than 7 , or $q=10$, and the 1st term, $p$, is 3 less than 7 , or $p=4$.

Thus, $p=4$ and $q=10$.

Solution 2

Suppose that the given arithmetic sequence has common difference $d$.

In terms of $p$ and $d$, the four terms are $p, p+d, p+2 d, p+3 d$.

Since the four terms are $p, 7, q, 13$, then $p+d=7$ and $p+3 d=13$.

Subtracting the first of these equations from the second, we obtain $2 d=6$ or $d=3$.

Since $p+d=7$ and $d=3$, then $p=4$.

Finally, $q=p+2 d=4+2(3)=10$.

Thus, $p=4$ and $q=10$.

(c) Solution 1

Moving from the 1st term to the 4th term involves adding the common difference 3 times. Since the difference between the 1st term and the 4th term is $(a+21)-a=21$, then 3 times the common difference equals 21 , which means that the common difference is 7 . The difference between the 3rd term, $c$, and the 1st term, $a$, is twice the common difference, or 14 , and so $c-a=14$.

## Solution 2

Suppose that the given arithmetic sequence has common difference $d$.

In terms of $a$ and $d$, the four terms are $a, a+d, a+2 d, a+3 d$.

Since the four terms are $a, b, c, a+21$, then $a+3 d=a+21$ or $3 d=21$, and so $d=7$.

Since $c=a+2 d$, then $c=a+14$.

Finally, $c-a=(a+14)-a=14$.

(d) Since $(y-6),(2 y+3),\left(y^{2}+2\right)$ form an arithmetic sequence in that order, then the differences between consecutive terms are equal and so

$$
\begin{aligned}
(2 y+3)-(y-6) & =\left(y^{2}+2\right)-(2 y+3) \\
y+9 & =y^{2}-2 y-1 \\
0 & =y^{2}-3 y-10 \\
0 & =(y-5)(y+2)
\end{aligned}
$$

Therefore, $y=5$ or $y=-2$.

We verify that each of these values of $y$ gives an arithmetic sequence.

When $y=5$, the sequence is $-1,13,27$ which is an arithmetic sequence with common difference 14 .

When $y=-2$, the sequence is $-8,-1,6$ which is an arithmetic sequence with common difference 7 .