Gee, I wish I could use my TI – 83!. For each of the following sequences, determine the common difference and the level at which it occurs. 1. -3, 0,

Gee, I wish I could use my TI – 83!

For each of the following sequences, determine the common difference and the level at which it occurs.

1. -3, 0, 5, 12, 21

2. 1, -2, -9, -20, -35

3. 6, 10, 14, 18,

4. - 10, -27, -56, -97

3 5 7 9

2 2 2

d = 2 at Level D2 Quadratic

-3 -7 -11 -15

-4 -4 -4 d = -4 at Level D2 Quadratic

4 4 4d = 4 at Level D1 Arithmetic

-17 -29 -41

-12 -12d = -12 at Level D2 Quadratic

Now use the function to generate the first four terms for each of these quadratic functions. Determine the common difference. What relation does it have with the coefficient of the a2 term?

132 xxy 124 2 xxy

xxy 26 22xy

Quadzilla

X Y

1 5

2 11

3 19

4 29

X Y

1 -5

2 -19

3 -41

4 -71

X Y

1 7

2 26

3 57

4 100

X Y

1 -2

2 -8

3 -18

4 -32

d is 2 at D2

a is 1d is -8 at D2

a is -4

d is 12 at D2

a is 6d is -4 at D2

a is -2

Is there a PATTERN here?

In a Quadratic Sequence there is a special relationship between a & d!

The Relationship between a & d in a Quadratic Sequence

The difference “d” from Level D2 is twice the coefficient of the n 2 or x 2 term in the general formula.

2

2

da

or

ad

So to determine the formula or rule for the nth term of a certain quadratic sequence, we must first find the common difference

and divide by 2 to find the coefficient “a”!

The Formula for the nth term of a Quadratic Sequence is

cbnantn 2

1. To algebraically determine the formula or expression for the nth term of a Quadratic Sequence we need to know the formula.

2. We need to know the common difference in order to determine the coefficient “a”.

7, 16, 29, 46, 67

3. We need to use the information from two terms to set up two equations.

7&171 ntnusgivest

16&2162 ntnusgivest

1

2

4. We need to solve the resulting System of Equations to determine “b” & “c”

5. We need to replace a, b, & c in the general formula.

STEPS

The Sequence 7, 16, 29, 46, 67

1 cbnantn 2

2 d = 4 so a = 2

3 Two terms to set up two equations.

cb

cb

cb

cbnant

tnt

n

n

5

27

)1()1(27

7&17

2

2

1

cb

cb

cb

cbnant

tnt

n

n

28

2816

)2()2(216

16&216

2

2

2

4 Solve for b & c by solving the System of Equations.

cb5cb28SUBTRACT

b3cb5

SUBTITUTE to find the other variable.

c 35c2

METHOD

232 2 nntn

Now we have

Exploration: Each diagram shows the number of line segments needed to connect a set of n points, no three of which lie in a straight line.

n=1 n=2 n=3 n=4

Create a Sequence with at least six terms to show the relationship between the number of points and the number of line segments needed to connect every point to every other point.

# of Points 1 2 3 4 5 6# of Line Segments

0 1 3 6

Exploration 2: Diagonals are formed on regular polygons starting with a three sided polygon or equilateral triangle. The number indicates the number of sides in the polygon.

n=3 n=4 n=5 n=6

Determine the number of diagonals in a regular polygon with 2- sides. Then determine the equation for the number of diagonals in a regular polygon of n sides.

# of sides 3 4 5 6 n# of

Diagonals0 1 3 6

Now You Try! Use “d” to determine “a” and then solve two equations!

The sequence is 7, 16, 31, 52, 79

9 15 21 27

6 6 6d2 =

a = 3 so cbnntn 23

Use t1 7 = 3(1)2 + b (1) +c

7 = 3 (1) +b +c

7 = 3 + b + c

4 = b + c

Use t2 16 = 3(2)2 + b (2) +c

16 = 3 (4) + 2b + c

16 = 12 + 2b + c

4 = 2b + c

4 = b + c

4 = 2b + c

0 = b

SUBSTITUTE 4 = b + c

4 = 0 + c

4 = c

43 2 ntn

Page 13 # 40, 41, 42

16 # 5, 8, 9

Remember, Homework is not meant to

be a burden. It is meant to help you to

reinforce the lesson and it helps you to

remember the steps and

proves whether you

understand!