Arithmetic sequences and series
Geometric sequences and series
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100

What is this formula used for?

t= a + (n - 1)d

Finding the formula for the general term (tn)

100

a) Define "geometric series"

b) What formula is used for finding a geometric series?

a) The sum of a geometric sequence

b)  Sn = [a(rn - 1)]/(r - 1)

100

Alex bought $7000 worth of rare coins. The coins appreciate in value by $200 a year. If he sells the coins for $11000, how many years ago did he buy them?

End of first year: t1

t1 = 7200

t2 = 7400 

t3 = 7600

      

    tn= a + (n - 1)d

   11000 = 7200 + (n - 1)200

11000 = 7200 + 200n - 200

11000 = 7000 + 200n

  4000 = 200n

      20 = n

.: Alex bought the coins 20 years ago

100

Given the explicit formula, f(n) = 2n – 3, write the first 5 terms

f(1) = 2(1) – 3                  f(2) = 2(2) – 3

   = -1                                = 1


f(3) = 2(3) – 3                  f(4) = 2(4) – 3 

   = 3                                 = 5


f(5) = 2(5) – 3

   = 7

100

How do you get consecutive rows of Pascal's Triangle?

By adding the two numbers above.

Ex. 

1

1    1

1    2    1

1    3    3    1

1    4    6     4    1

200

Find the formula for the general term (tn) and for t22 for the sequence: 4, 9, 14,...

a = 4     d = 5

t= 4 + (n - 1)5

   = 4 + 5n - 5

   = 5n - 1


t22 = 5(22) -1

     =109

200

Determine the formula for the nth term, tn, and find t5 for the geometric sequence    54, 27, 13.5, ...

a = 54         r = 1/2


tn = arn - 1

tformula: tn = 54(1/2)n - 1


t5 = 54(1/2)5 - 1

     = 54(1/2)4

     = 3.375

200

A certain bacteria has a population of 24 at the beginning of an experiment and doubles every minute. If the experiment lasted 2 hours, how many bacteria were there in the end?

a = 24   r = 2

2 hours = 60 minutes x 2 = 120


Sn = n/2 [(2a + (n - 1)d]

S120 = 120/2 [(2(24) + (120 - 1)2]

       = 60[48 + 119(2)]

       = 60(286)

       = 17,160

.: There were 17,160 bacteria at the end of the experiment.

200

Write the first five terms

t1 = 4      tn = tn-1(3)

t= t2-1(3)          t= t3-1(3)         t= t4-1(3) 

   = 4(3)                 = 12(3)             = 36(3)

   = 12                   = 36                  = 108


t= t5-1(3)

   = 108(3)

   = 324

200

Using the sequence      5, 12, 19, 26, 33

a) Determine the finite difference

b) Determine the explicit formula for the nth term

n = term number        tn = term value

tn: 5  12  19   26  33

n:  1   2    3    4    5

The finite difference is 7.

b) 7 must be in our formula since it is the common difference. If we multiply 7 by each term number, the result is 2 more than the t. Hence, the explicit formula is:

tn = 7n - 2

300

Find the sum of the first 45 terms of the series

5 + 11 + 17 + ...

 a = 5    d = 6    n = 45

Sn = n/2 [2a +(n - 1)d]

S45 = 45/2 [2(5) +(45 - 1)6]

     = 22.5[10 +(44)6]

     = 22.5(10 + 264)

     = 22.5(274)

     = 6165

300

Find S10 for the series 3 + 7 + 11 + …

S10 = [3(410 – 1)]/4 – 1

      = [3(1048,576)]/3

      =1048,576

300

In an arithmetic sequence t3 = 16 and t17= 128. Determine the first term, the common difference, and the sum of the first 20 terms.

d = (128-16)/(17-3)        tn= a + (n - 1)d

   = 112/14                   16 = a + (3 - 1)8

   = 8                            16 = a + (2)8

                                    16 = a + 16

                                      0 = a

Sn = n/2 [2a +(n - 1)d]

S20 = 20/2 [2(0) +(20 - 1)8]

     = 10[(19)8]

     = 10(152)

     = 1520

300

Expand using Pascal's Triangle

(3x + 9)5

1(3x)5(9)+ 5(3x)4(9)1 + 10(3x)3(9)+ 10(3x)2(9)3 + 5(3x)1(9)4 + 1(3x)0(9)5

= 243x5 + 3645x4 + 21870x3 + 65610x2 + 98415x + 59049

300

Expand using Pascal's Triangle

(7x–4)4

=1(7x)4(4)0 + 4(7x)3(4)1 – 6(7x)2(4)2 + 4(7x)1(4)3 – 1(7x)0(4)4

=2401x4 + 5488x3 – 4704x2 + 1792x – 256

400

How many terms are there in the sequence 1, 3, 6, ..., 57?

t= a + (n - 1)d

57 = 1 + (n - 1)3

57 = 1 + 3n - 3

57 = 3n - 2

59 = 3n

19= n

.: There are 19 terms in the sequence

400

Find the number of terms in the geometric sequence 

2, 6, 18, ..., 486

a = 2      r = 3


tn = arn - 1

486 = 2(3)n - 1

486/2 = (2/2)(3)n - 1

243 = 3n - 1

 35 = 3n - 1


n - 1 = 5

n = 6

.: There are 6 terms in the sequence.

400

A new restaurant sold 10 four-course meals on their opening day. Each day after, the number of meals they sold increased by 5 from the day before.

a) Write the formula for the nth term of the sequence.

b) How many meals and how many courses were sold 1 week after opening?

t1 = 10   t2 = 15     t3 = 20     a = 10    d = 5


a)  tn= a + (n - 1)d 

   tn= 10 + (n - 1)5

      = 10 + 5n - 5

      = 5 + 5n


b) Sn = n/2 [2a + (n - 1)d]

    S7 = 7/2 [2(10) + (7 - 1)5]

        = 3.5[20 + (6)5]

        = 3.5(40)

        = 140

140 meals x 4 courses = 560 courses

.: The restaurant sold 140 meals with a total of 560 courses.

400

Describe the pattern in this sequence and write the next 3 terms:

1, 1, 2, 3, 5, 8, 13, 21, ...

tn is the sum of the last two terms

34, 55, 89

400

Describe the pattern in this sequence and write the next three terms.

38, 8, 2, ...

The rate is 1/4.

0.5, 0.125, 0.03125

500

Find the sum of the arithmetic series given the first and last terms:  a = 2     t10 = 74

d = (74 - 2)/(10 - 1)

   = 72/9

   = 8


Sn = n/2 [2a +(n - 1)d]

S10 = 10/2 [2(2) +(10 - 1)8]

      = 5[4 + 9(8)]

      = 5(76)

      = 380

500

Find the sum of the series 2 + 8 + 32 + … + 512

r = 4      a = 2


tn = arn - 1

512 = 2(4)n - 1

512/4 = 2(4/4)n - 1

128 = 2n - 1

  27 = 2n - 1


n - 1 = 7

     n = 8


Sn = [a(rn - 1)]/r - 1

S8 = [2(48 - 1)]/4 - 1

    = [2(65,535)]/3

    = 43690.7

500

A record-breaking snowstorm in Yukon resulted in 40 cm of snow. However, the temperature rose and one-sixth of the snow melted daily. How much snow was left after 1 week?

If ⅙ is melting, ⅚ are remaining.

a = 40 x 5/6 = 33.3     d = 5/6


tn= arn – 1

t7= 33.3(5/6)7 – 1

t7= 33.3(5/6)6

t7= 11.2

.: There will be 11.2 cm of snow left after 1 week.

500

Determine a recursion formula for this sequence and find the 5th term

3, 8, 13, 18, ...

a=3        d=5

Since the common difference is 5, the nth term will always be the previous term + 5. Therefore, the formula is: 

t1 = 3,    t= tn-1 + 5

t5= t5-1 + 5

    = t4 + 5

    = 18 + 5

    = 23

500

Determine the formula for calculating numbers in Fibonacci's Sequence.

1, 1, 2, 3, 5, 8, 13, 21, ...

Since each term is the sum of the previous two terms, the formula must be:

tn = tn -1 + tn - 2

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