Arithmetic

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# Arithmetic - PowerPoint PPT Presentation

Arithmetic. Sequences &amp; Series jeff.bivin@lz95.org. Last Updated: October 11, 2005. Arithmetic Progression. n th term. 5, 8, 11, 14, 17, 20, … 3n+2, … -4, 1, 6, 11, 16, … 5n – 9, . . . 11, 7, 3, -1, -5, … -4n + 15, . . . n th term. Jeff Bivin -- LZHS.

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### Arithmetic

Sequences & Series

jeff.bivin@lz95.org

Last Updated: October 11, 2005

Arithmetic Progression

nth term

5, 8, 11, 14, 17, 20, … 3n+2, …

-4, 1, 6, 11, 16, … 5n – 9, . . .

11, 7, 3, -1, -5, … -4n + 15, . . .

nth term

Jeff Bivin -- LZHS

nth term of arithmetic sequence

Tn = a + d(n – 1)

a = First term

d = common difference

n = number of terms.

Common difference = the difference between two consecutive terms in a sequence.

d = Tn – Tn-1

Jeff Bivin -- LZHS

Find the nth term of the following AP.

8, 11, 14, 17, 20, …

First term is 8

Common difference is 3

Tn = a + d(n – 1)

Tn = 8 + 3(n – 1)

Tn = 8 + 3n – 3

Tn = 3n + 5

Jeff Bivin -- LZHS

Finding the nth term

6, 1, 8, 15, 22, …

First term is -6

common difference is 7

Tn = a + d(n – 1)

Tn = -6 + 7(n – 1)

Tn = -6 + 7n – 7

Tn = 7n - 13

Jeff Bivin -- LZHS

Finding the nth term

23, 19, 15, 11, 7, …

First term is 23

common difference is -4

Tn = a + d(n – 1)

Tn = 23 + -4(n – 1)

Tn = 23 -4n + 4

Tn = -4n + 27

Jeff Bivin -- LZHS

Finding the 956th term

a1 = 156

d = -16

n = 956

156, 140, 124, 108, . . .

Tn = a + d(n – 1)

T956 = 156 + -16(956 – 1)

T956 = 156 - 16(955)

T956 = 156 - 15280

T956 = -15124

Jeff Bivin -- LZHS

Finding the 100th term

a = 5

d = 6

n = 100

5, 11, 17, 23, 29, . . .

Tn = a + d(n – 1)

T100 = 5 + 6(100 – 1)

T100 = 5 + 6(99)

T100 = 5 + 594

T100 = 599

Jeff Bivin -- LZHS

Finding the number of terms in the AP

a = 10

d = -2

Tn = -24

10, 8, 6, 4, 2, . . .-24

Tn = a + d(n – 1)

-24 = 10 -2(n – 1)

-34 = -2(n – 1)

17 = n-1

n = 18

Jeff Bivin -- LZHS

The 5th term of an AP is 13 and the 13th term is -19. Find the first term & the common difference.

T5 = a + 4d = 13……..(1)

T13= a + 12d = -19……….(2)

(2) – (1): 8d = -19 - 13

8d = - 32

d = -4

Substitute d = -4 into (1):

a + 4(-4) = 13

a – 16 = 13

a = 29

Jeff Bivin -- LZHS

Problem solving

In a race, a swimmer takes 35 seconds to

swim the first 100 m, 39 seconds to swim

the second 100m and 43 seconds to swim

the third 100 m. If he continues to swim in

this manner, how long does he take to finish

the 10th lap of 100m.

Summing it up

Sn = a1 + (a1 + d) + (a1 + 2d) + …+ an

Sn = an + (an - d) + (an - 2d) + …+ a1

Jeff Bivin -- LZHS

1 + 4 + 7 + 10 + 13 + 16 + 19

a1 = 1

an = 19

n = 7

Jeff Bivin -- LZHS

4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24

a1 = 4

an = 24

n = 11

Jeff Bivin -- LZHS

Find the sum of the integers from 1 to 100

a1 = 1

an = 100

n = 100

Jeff Bivin -- LZHS

Find the sum of the multiples of 3 between 9 and 1344

Sn = 9 + 12 + 15 + . . . + 1344

a1 = 9

an = 1344

d = 3

Jeff Bivin -- LZHS

Find the sum of the multiples of 7 between 25 and 989

Sn = 28 + 35 + 42 + . . . + 987

a1 = 28

an = 987

d = 7

Jeff Bivin -- LZHS

Evaluate

Sn = 16 + 19 + 22 + . . . + 82

a1 = 16

an = 82

d = 3

n = 23

Jeff Bivin -- LZHS

Evaluate

Sn = -29 - 31 - 33 + . . . - 199

a1 = -29

an = -199

d = -2

n = 86

Jeff Bivin -- LZHS

Sn = 10 01+ 1012 + 1023 + ... + 9999

a1 = 1001

an = 9999

d = 11

Jeff Bivin -- LZHS

Review -- Arithmetic

Sum of n terms

nth term

Jeff Bivin -- LZHS