# 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 \\ &= \boxed{92} \quad \quad \textbf{(A)}\\ \end{align*}

- $\text{herobrine-india}$

## Solution 4

In an actual contest, this would probably take too much time but is nevertheless a solution. The general formula for computing sums of any arithmetic sequence where $x$ is the number of terms, $f$ is the first term and $l$ is the last term is $\frac{(f+l)x}{2}$. If one uses that formula for both sequences, they will get $2530$ as the sum for $X$ and $2622$ as the sum for $Y$. Subtracting $X$ from $Y$, one will get the answer $\boxed{92 \text{\textbf{ (A)}}}$. - danfan

~savannahsolver