# Difference between revisions of "2011 AMC 10A Problems/Problem 4"

## Problem

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 $Y - X?$ $\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$

## Solution 1

We see that both sequences have equal numbers of terms, so reformat the sequence to look like: \begin{align*} Y = \ &12 + 14 + \cdots + 100 + 102\\ X = 10 \ + \ &12 + 14 + \cdots + 100\\ \end{align*} From here it is obvious that $Y - X = 102 - 10 = \boxed{92 \ \mathbf{(A)}}$.

### Note

Another way to see this is to let the sum $12+14+16+...+100=x.$ So, the sequences become \begin{align*} X = 10+x \\ Y= x+102 \\ \end{align*}

Like before, the difference between the two sequences is $Y-X=102-12=92.$

## Solution 2

We see that every number in Y's sequence is two more than every corresponding number in X's sequence. Since there are 46 numbers in each sequence, the difference must be: $46\cdot 2=\boxed{92}$

## Solution 3 \begin{align*} X&=10+12+14+\cdots +100 \\ Y&=X-10+102 = X+92 \\ Y-X &= (X+92)-X \\ Y-X &= X-X+92 \\ Y-X &= 0+92 \\ Y-X &= \boxed{92} \quad \quad \textbf{(A)}\\ \end{align*} $\blacksquare$

- $\text{herobrine-india}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 