1. A sequence of three numbers $a, b, c$ form an arithmetic sequence if the difference between successive terms in the sequence is the same. That is, when $b-a=c-b$.
(a) The sequence $2, b, 8$ forms an arithmetic sequence. Determine $b$.
(b) Given a sequence $a, b, c$, let $d_{1}$ be the non-negative number to increase or decrease $b$ by so that, without changing $a$ or $c$, the result is an arithmetic sequence. Let $d_{2}$ be the positive number to increase or decrease $c$ by so that, without changing $a$ or $b$, the result is an arithmetic sequence.
For example, if the three-term sequence is $3,10,13$, then we need to decrease 10 to 8 to make the arithmetic sequence $3,8,13$. We decreased $b$ by 2 , so $d_{1}=2$. If we change the third term, we need to increase 13 to 17 to make the arithmetic sequence $3,10,17$. We increased 13 by 4 , so $d_{2}=4$.
Suppose the original three term sequence is $1,13,17$. Determine $d_{1}$ and $d_{2}$.
(c) Define $d_{1}, d_{2}$ as in part (b). For all three-term sequences, prove that $2 d_{1}=d_{2}$.
(a) Solution 1: We require that $b-a=c-b$, so $b-2=8-b$ and $b=5$.
Solution 2: $b$ must be in the middle of $a$ and $c$, so $\mathrm{b}=(8+2) / 2=5$.
Solution 3: $2,5,8$ forms an arithmetic sequence, so $b=5$.
(b) Solution 1: $b-a=c-b$, and $a=1, c=17$. Substituting into the equation and solving gives $b=9$. Since we have 13 , as our middle term, the difference must be $d_{1}=13-9=4$.
Similarly, if we have $a=1, b=13$, solving gives $c=25$. The difference must be $d_{2}=$ $25-17=8$.
Solution 2: The second number must be 9 to form an arithmetic sequence, so $d_{1}=$ $13-9=4$.
Similarly, the third number must be 25 to form an arithmetic sequence, so $d_{2}=25-17=$ 8 .
(c) Solution 1: We can rewrite the relation for an arithmetic sequence as $2 b=a+c$. Suppose we have a sequence (not necessarily arithmetic) $a, b, c$, then we can write $2\left(b \pm d_{1}\right)=a+c$, and we can also write $2 b=a+\left(c \pm d_{2}\right)$. Subtracting one equation from the other we are left with $2 d_{1}= \pm d_{2}$ and since both $d_{1}$ and $d_{2}$ are positive $2 d_{1}=d_{2}$.
Solution 2: We first determine $d_{1}$. Let $b^{\prime}$ be the number we replace $b$ with to form an
arithmetic sequence. Then $b^{\prime}-a=c-b^{\prime}$. Therefore, $b^{\prime}=(a+c) / 2$. Hence, we changed $b$ by $(a+c) / 2-b$. Hence, $d_{1}=|(a+c) / 2-b|=|(a+c-2 b) / 2|$, where $|\cdot|$ denotes absolute value. (The contestant shouldn't need to use absolute value. Any explanation that we take the positive version of the number would suffice.)

We next determine $d_{2}$. Let $c^{\prime}$ be the number to replace $c$ by to make the sequence arithmetic. Then $c^{\prime}-b=b-a$. Therefore, $c^{\prime}=2 b-a$. Hence, we change $c$ by $c^{\prime}-c=2 b-a-c$. Therefore, $d_{2}=|2 b-a-c|=2|(2 b-a-c) / 2|=2|(a+c-2 b) / 2|=2 d_{1}$, as desired.

Solution 3: Place the points $A(1, a), B(2, b), C(3, c)$ on the coordinate plane. Let $b^{\prime}$ be the number so that $a, b^{\prime}, c$ is an arithmetic sequence and $c^{\prime}$ the number so that $a, b, c^{\prime}$ is an arithmetic sequence. Let $B^{\prime}=\left(2, b^{\prime}\right)$ and $C^{\prime}=\left(3, c^{\prime}\right)$. Then $A, B^{\prime}, C$ lie on a straight line and $A, B, C^{\prime}$ lie on a straight line. Furthermore, $B^{\prime}$ is the midpoint of $A C$ and $B$ is the midpoint of $A C^{\prime}$.

Note that $d_{1}$ is the distance between $B$ and $B^{\prime}$ and $d_{2}$ is the distance between $C$ and $C^{\prime}$. Consider triangle $A C C^{\prime}$. Since $B B^{\prime}, C C^{\prime}$ are both vertical, they are parallel. Therefore, $\triangle A B B^{\prime}$ and $\triangle A C^{\prime} C$ are similar, with ratio $1: 2$. Therefore, $C C^{\prime}$ is twice the length of $B B^{\prime}$, yielding $d_{2}=2 d_{1}$.