AMC 10 · 2003 · #24
Grade 8 algebraThe first four terms in an arithmetic sequence are x+y, x−y, xy, and yx, in that order. What is the fifth term?
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Four expressions — $x+y$, $x-y$, $xy$, and $\frac{x}{y}$ — form the first four terms of an arithmetic sequence, in that order. In an arithmetic sequence every step from one term to the next adds the same fixed amount. Find the fifth term.
Givens: The first four terms, in order, are $x+y$, $x-y$, $xy$, $\frac{x}{y}$.; The four terms form an arithmetic sequence, so consecutive terms differ by a constant.; Answer choices: (A) $-\frac{15}{8}$, (B) $-\frac{6}{5}$, (C) $0$, (D) $\frac{27}{20}$, (E) $\frac{123}{40}$.
Unknowns: The fifth term of the sequence.; The values of $x$ and $y$ that make the four given expressions arithmetic.
Understand
Restated: Four expressions — $x+y$, $x-y$, $xy$, and $\frac{x}{y}$ — form the first four terms of an arithmetic sequence, in that order. In an arithmetic sequence every step from one term to the next adds the same fixed amount. Find the fifth term.
Givens: The first four terms, in order, are $x+y$, $x-y$, $xy$, $\frac{x}{y}$.; The four terms form an arithmetic sequence, so consecutive terms differ by a constant.; Answer choices: (A) $-\frac{15}{8}$, (B) $-\frac{6}{5}$, (C) $0$, (D) $\frac{27}{20}$, (E) $\frac{123}{40}$.
Plan
Primary tool: #13 Convert to Algebra
Secondary: #4 Introduce a Variable, #5 Look for a Pattern, #7 Identify Subproblems
The only fact we are given is a pattern — 'arithmetic sequence' — so first turn that pattern into arithmetic (Tool #5, Tool #13): equal gaps mean each term minus the one before it equals the same constant $d$. The first two terms are simple, so read off $d$ from them (Tool #4, name the gap $d$). That single value $d$ then lets us write the third and fourth terms two ways — as the given expressions, and as 'previous term plus $d$'. Setting those equal gives two equations in $x$ and $y$. Solving that little system is the core subproblem (Tool #7); once $x$, $y$, and $d$ are known, the fifth term is just the fourth term plus $d$.
Execute — Answer: E
7.EE.A.1 Step 1 Read the common gap off the first two terms
- In an arithmetic sequence the gap $d$ between neighbors is constant, and the first two terms are the easiest place to find it.
- Subtract the first term from the second: $d = (x-y) - (x+y)$.
- The $x$'s cancel and the $-y$ combines with the $-y$, leaving $d = -2y$.
- So every step in this sequence adds $-2y$.
💡 The steady gap must equal the easiest difference we can compute, and that is second-term minus first-term.
6.EE.B.6 Step 2 Write the third and fourth terms two ways
- Each term is 'the term before it, plus $d = -2y$'.
- The third term is the second term plus $d$: $(x-y) + (-2y) = x - 3y$.
- But the third term is also given as $xy$, so $x - 3y = xy$.
- The fourth term is the third term plus $d$: $xy + (-2y) = xy - 2y$.
- It is also given as $\frac{x}{y}$, so $xy - 2y = \frac{x}{y}$.
- Two equations now hold at once.
💡 Anything that equals a term two different ways gives an equation — set the given expression equal to 'previous term plus the gap.'
8.EE.C.7 Step 3 Eliminate x to get an equation in y
- Equation (1) says $xy = x - 3y$.
- Use that to remove the $xy$ inside equation (2).
- Replace $xy$ in (2) with $x - 3y$: $(x - 3y) - 2y = \frac{x}{y}$, which tidies to $x - 5y = \frac{x}{y}$.
- Multiply both sides by $y$ to clear the fraction: $xy - 5y^2 = x$.
- Now substitute $xy = x - 3y$ one more time: $(x - 3y) - 5y^2 = x$.
- The $x$ on each side cancels, leaving $-3y - 5y^2 = 0$, that is $y(5y + 3) = 0$.
💡 Feeding one equation into the other kills the mixed $xy$ and $x$ terms, collapsing everything down to a single equation in $y$.
7.NS.A.2 Step 4 Solve for y, then x, then the gap
- From $y(5y+3)=0$ the two possibilities are $y = 0$ or $y = -\frac{3}{5}$.
- But $y = 0$ is impossible: the fourth term $\frac{x}{y}$ would divide by zero.
- So $y = -\frac{3}{5}$.
- Put this into equation (1), $x - 3y = xy$, written as $x(1 - y) = 3y$: with $1 - y = 1 + \frac{3}{5} = \frac{8}{5}$ and $3y = -\frac{9}{5}$, we get $x = \frac{-9/5}{8/5} = -\frac{9}{8}$.
- The gap is $d = -2y = -2\left(-\frac{3}{5}\right) = \frac{6}{5}$.
💡 Dividing by $y$ was only allowed because $y\neq 0$, which is exactly why the $y=0$ root gets thrown out.
7.NS.A.1 Step 5 Add the gap to the fourth term
- The fifth term is the fourth term plus $d$.
- The fourth term is $\frac{x}{y} = \frac{-9/8}{-3/5} = \frac{9}{8}\cdot\frac{5}{3} = \frac{15}{8}$.
- Add $d = \frac{6}{5}$: put both over $40$, so $\frac{15}{8} = \frac{75}{40}$ and $\frac{6}{5} = \frac{48}{40}$, giving $\frac{75}{40} + \frac{48}{40} = \frac{123}{40}$.
- That is choice (E).
💡 Once the constant gap is known, the next term is always just the last term plus that gap.
7.EE.A.1 In an arithmetic sequence the gap $d$ between neighbors is constant, and the fir 6.EE.B.6 Each term is 'the term before it, plus $d = -2y$'. The third term is the second 8.EE.C.7 Equation (1) says $xy = x - 3y$. Use that to remove the $xy$ inside equation (2) 7.NS.A.2 From $y(5y+3)=0$ the two possibilities are $y = 0$ or $y = -\frac{3}{5}$. But $y 7.NS.A.1 The fifth term is the fourth term plus $d$. The fourth term is $\frac{x}{y} = \f Review
Reasonableness: Check that the four given expressions really are arithmetic with $x=-\frac{9}{8}$, $y=-\frac{3}{5}$. In fortieths: $x+y = -\frac{45}{40}-\frac{24}{40} = -\frac{69}{40}$; $x-y = -\frac{45}{40}+\frac{24}{40} = -\frac{21}{40}$; $xy = \frac{27}{40}$; $\frac{x}{y} = \frac{75}{40}$. The differences are $-\frac{21}{40}-(-\frac{69}{40}) = \frac{48}{40}$, $\frac{27}{40}-(-\frac{21}{40}) = \frac{48}{40}$, and $\frac{75}{40}-\frac{27}{40} = \frac{48}{40}$ — all equal to $\frac{48}{40}=\frac{6}{5}=d$. The sequence is genuinely arithmetic, so extending by one more $\frac{48}{40}$ gives $\frac{123}{40}$, matching (E).
Alternative: Instead of eliminating $x$, solve equation (1) for $x$ directly: $x - 3y = xy$ gives $x(1-y) = 3y$, so $x = \frac{3y}{1-y}$. Substitute this into equation (2) and clear denominators to get a single equation in $y$; it reduces to the same $5y + 3 = 0$, so $y = -\frac{3}{5}$ and $x = -\frac{9}{8}$ as before. Either order of substitution lands on the same values and the same fifth term.
CCSS standards used (min grade 8)
7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Simplifying $(x-y)-(x+y)$ to $-2y$ to read the common difference off the first two terms.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Turning 'each term is the previous term plus $d$' into the two equations $x-3y=xy$ and $xy-2y=\frac{x}{y}$.)8.EE.C.7Solve linear equations in one variable (Substituting one equation into the other to eliminate $x$ and reduce to $y(5y+3)=0$.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Solving $x(1-y)=3y$ for $x=-\frac{9}{8}$ and computing the gap $d=-2y=\frac{6}{5}$.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Adding the fourth term $\frac{15}{8}$ and the gap $\frac{6}{5}$ over a common denominator to reach $\frac{123}{40}$.)
⭐ Turn 'arithmetic sequence' into 'each term equals the one before plus the same gap' — that gives you equations to pin down the unknowns, and then the next term is just add the gap once more.
⭐ Turn 'arithmetic sequence' into 'each term equals the one before plus the same gap' — that gives you equations to pin down the unknowns, and then the next term is just add the gap once more.
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