Sequences and equations

Question

 1.    For the design below: 

         http://www.flickr.com/photos/99973477@N03/11444159853/

 

a)    Draw the next three shapes in the sequence.

b)    Draw up a table for the first ten shapes in the sequence, and the number of matches required to complete the shape.

c)    Determine the word relationship for the sequence of outputs.

d)    Determine the formula which would allow you to find the number of matches in any shape in the sequence.

e)      Draw a graph of the relationship.

 

2.       For the design below:

          http://www.flickr.com/photos/99973477@N03/11444055415/

a)  Draw the next three shapes in the sequence.

b)  Draw up a table for the first ten shapes in the sequence, and the number of matches required to complete the shape.

c)  Determine the word relationship for the sequence of outputs.

d) Determine the formula which would allow you to find the number of matches in any shape in the sequence.

e) Draw a graph of the relationship.

 

3. A student banks $200 when she starts secondary College in Grade 7. In Grade 8 she banks $250,$300 in Grade 9, and so on. She is at school for 6 years.

 

a)  Determine a formula relating the amount deposited during a year (200,250,...) to how many years they have been saving(1,2,...).

 

Because the deposits are increasing, the formula for the bank balance is more complex and is not a linear relationship. Answer the following using a table.

 

b)  What is the total amount she has banked by the time she finishes after 6 years?

c)   If she continued banking in this manner after she leaves school, how many years(from the start) will it take for her balance to exceed $15,000?

Answer 1

b) she banked by the time she finishes after 6 years total amont is $1950.

c) If she continued banking in this manner after she leavesa school,

22years will take for her balance to exceed $15000.

Answer 2

(1).

(a). The next three shapes in the sequence :

 

(b). Make the table

Numbers	1	2	3	4	5	6	7	8	9	10
Number of shapes	1	2	3	4	5	6	7	8	9	10

 

(c). The word relationship for the sequence of outputs :

Number of shapes = number(n).

 

(d). Determine the formula which would allow you to find the number of matches in any shape in the sequence.

Sequence : 1, 2, 3, 4, . . . . . . .n.

Since common difference is 1, the above sequence is arithmetic sequence.

nth term t_n = a₁ + (n - 1)d, where a₁ = first term and d = common term.

t_n = 1 + (n - 1)1

t_n = 1 + n - 1

t_n = n.

 

(e). Draw a graph of the relationship y = x.

 

Answer 3

(2).

(a). The next three shapes in the sequence :

 

(b). Make the table

Numbers	1	2	3	4	5	6	7	8	9	10
Number of shapes	1	4	7	10	13	16	19	22	25	28

 

(c). The word relationship for the sequence of outputs :

Number of shapes = The difference between three times a number(n) and two.

 

(d). Determine the formula which would allow you to find the number of matches in any shape in the sequence.

Sequence : 1, 4, 7, 10, . . . . . . .nth term.

Since common difference is 3, the above sequence is arithmetic sequence.

nth term t_n = a₁ + (n - 1)d, where a₁ = first term and d = common term.

t_n = ₁ + (n - 1)3

t_n = 1 + 3n - 3

t_n = 3n - 2.

 

(e). Draw a graph of the relationship y = 3x - 2.

Answer 4

(3).(a).

Year                     Balance after each compoundig.

  0                        200                            [P = P]

  1                        200 + 250 = 450        [P₁ = P + Pr = P(1 + r)]

  2                        450 + 300 = 750        [P₂ = P₁(1 + r) = P(1 + r)(1 + r) = P(1 + r)²]

  3                        750 + 350 = 1100      [P₃ = P₂(1 + r) = P(1 + r)²(1 + r) = P(1 + r)³]

  .                        .                                             .

  .                        .                                             .

  .                        .                                             .

  t                        .                                [P_t = P(1 + r)^t]

For n compounings per year : A = P(1 + r/n)^nt, where P = principal, r = interest rate(in decimal form), t = number of years.

For continuous compouning : A = Pe^rt.

 

(b). The total amount after 6 years :

Here find the r value :

If A = 450, P = 200 and t = 1 then 450 = 200e^(1)r    ------> (1).

If A = 750, P = 200 and t = 2 then 750 = 200e^(2)r    ------> (2).

Divide the equation 2 by equation 1.

750/450 = 200e^2r / 200e^r -----> e^r = 1.67  -----> r = ln 1.67 = 0.5.

The total amount after 6 years A = (200)e^(0.5)(6) = 200 * 20.0855 = 4017.107 dollars.

 

(c). Find the time in years :

A = 15, 000 dollars and r = 0.5, P = 200.

A = Pe^rt.

15,000 = (200)e^(0.5)(t)

75 = e^t/2 =

ln 75 = t/2

8.635 years = t.