# 2005 AMC 10A Problems/Problem 17

## Problem

In the five-sided star shown, the letters $A, B, C, D,$ and $E$ are replaced by the numbers $3, 5, 6, 7,$ and $9$, although not necessarily in this order. The sums of the numbers at the ends of the line segments $AB$, $BC$, $CD$, $DE$, and $EA$ form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence? $\textbf{(A) } 9\qquad \textbf{(B) } 10\qquad \textbf{(C) } 11\qquad \textbf{(D) } 12\qquad \textbf{(E) } 13$

## Solution 1

Each corner $(A,B,C,D,E)$ goes to two sides/numbers. ( $A$ goes to $AE$ and $AB$, $D$ goes to $DC$ and $DE$). The sum of every term is equal to $2(3+5+6+7+9)=60$

Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is $\frac{60}{5}=\boxed{\textbf{(D) }12}$

## Solution 2

We know that the smallest number in this sequence must be $3 + 5 = 8$, and the biggest number must be $7 + 9 = 16$. Since there are $5$ terms in this sequence, we know that $8 + 4d = 16$, or that $d = 2$. Thus, we know that the middle term must be $8 + 2 \cdot 2 = \boxed{12}.$ ~yk2007 (Daniel K.)

## Video Solution by OmegaLearn

~ pi_is_3.14

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