Recursive and Explicit Formulas from: category_eng

Suppose that ${a_n}$ is an arithmetic sequence with $a_1+a_2+cdots+a_{100}=100 ext{ and } a_{101}+a_{102}+cdots+a_{200}=200.$ What is the value of $a_2 - a_1 ?$

$mathrm{(A) } 0.0001qquad mathrm{(B) } 0.001qquad mathrm{(C) } 0.01qquad mathrm{(D) } 0.1qquad mathrm{(E) }...$

The ï¬rst four terms in an arithmetic sequence are $x+y$ , $x-y$ , $xy$ , and $frac{x}{y}$ , in that order. What is the ï¬fth term?

$extbf{(A)} -frac{15}{8}qquad extbf{(B)} -frac{6}{5}qquad extbf{(C)} 0qquad extbf{(D)} frac{27}{20}qquad extb...$

A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains 100 cans, how many rows does it contain?

$(mathrm {A}) 5 qquad (mathrm {B}) 8 qquad (mathrm {C}) 9 qquad (mathrm {D}) 10 qquad (mathrm {E}) 11$

1.

Adding the two given equations together gives

$a_1+a_2+...+a_{200}=300$ .

Now, let the common difference be $d$ . Notice that $a_2-a_1=d$ , so we merely need to find $d$ to get the answer. The formula for an arithmetic sum is

$frac{n}{2}(2a_1+d(n-1))$ ,

where $a_1$ is the first term, $n$ is the number of terms, and $d$ is the common difference. Now we use this formula to find a closed form for the first given equation and the sum of the given equations. For the first equation, we have $n=100$ . Therefore, we have

$50(2a_1+99d)=100$ ,

$2a_1+99d=2$ . *(1)

For the sum of the equations (shown at the beginning of the solution) we have $n=200$ , so

$100(2a_1+199d)=300$

$2a_1+199d=3$ *(2)

Now we have a system of equations in terms of $a_1$ and $d$ .
Subtracting (1) from (2) eliminates $a_1$ , yielding $100d=1$ , and $d=a_2-a_1=frac{1}{100}=.01, oxed{ ext{C}}$ .

The difference between consecutive terms is $(x-y)-(x+y)=-2y.$ Therefore we can also express the third and fourth terms as $x-3y$ and $x-5y.$ Then we can set them equal to $xy$ and $frac{x}{y}$ because they are the same thing.

$egin{align*}xy&=x-3yxy-x&=-3yx(y-1)&=-3yx&=frac{-3y}{y-1}end{align*}$

Substitute into our other equation.

$egin{align*}frac{x}{y}&=x-5yfrac{-3}{y-1}&=frac{-3y}{y-1}-5y-3&=-3y-5y(y-1)�&=5y^2-2y-3�&=(5...$

But $y$ cannot be $1$ because then every term would be equal to $x.$ Therefore $y=-frac35.$ Substituting the value for $y$ into any of the equations, we get $x=-frac98.$ Finally,

$frac{x}{y}-2y=frac{9cdot 5}{8cdot 3}+frac{6}{5}=oxed{ extbf{(E)} frac{123}{40}}$

4.

The sum of the first $n$ odd numbers is $n^2$ . As in our case $n^2=100$ , we have $n=oxed{10}Longrightarrowmathrm{(D)}$ .

5. 59

6.	notSure