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2011 AMC 10A Problem 4

2011 AMC 10A Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 1 of 5 · about 2 min · position P1-10

Let $X$ and $Y$ be the following sums of arithmetic sequences: \begin{eqnarray} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{eqnarray} What is the value of $Y - X$ ?

A)
$92$
B)
$98$
C)
$100$
D)
$102$
E)
$112$

Answer

Key insight

The two sums share every term from 12 to 100, so Y minus X is just the leftover 102 minus the leftover 10.

Solution

Both sums run through the even numbers $12, 14, \dots, 100$ . Beyond that, $X$ has the extra term $10$ at the front and $Y$ has the extra term $102$ at the end.

When we subtract, every shared term cancels:
$Y - X = 102 - 10 = 92.$

The answer is $\boxed{\textbf{(A)}\ 92}$ .

Why this works

Two sums that overlap almost completely should be compared by cancellation, not by evaluating each one. Line up what is common and subtract only what differs. This is the same idea behind telescoping sums: the bulk of the work vanishes.

Alternative approach

Pair the terms in order: $12 - 10$ , $14 - 12$ , and so on up to $102 - 100$ . Each pair contributes $2$ , and there are $\frac{100 - 10}{2} + 1 = 46$ terms in each sum, so $Y - X = 2 \cdot 46 = 92$ .

The trap

Counting the terms wrong (45 or 47 instead of 46) when pairing term by term, or computing 102 - 100 = 2.

Common mistakes

Counting the terms wrong (45 or 47 instead of 46) when pairing term by term, or computing 102 - 100 = 2.
Computing both sums with the arithmetic-series formula and making an arithmetic slip, when almost everything cancels.

Techniques

Set up the equation/formula and compute; no special trick needed · Collapse a sum or product by cancellation