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Unit 8 Project

1. Find the next five terms of the arithmetic sequence 42, 37, 32β¦

Solution: the next five terms are: 27, 22, 17, 12, and 7

2. Find the 24th term of an arithmetic sequence for which a1 = 2 and d = 6

Solution: ππ = π1 + (π β 1)π = 2 + (24 β 1) Γ 6 = πππ

3. Find the three arithmetic means between -4 and 16

Solution:

Assume the arithmetic means to be: a, b, c, so the arithmetic sequence will

be: -4, a, b, c, 16

π1 = β4, π5 = 16

ππ = π1 + (π β 1)π

β΄ π5 = π1 + (5 β 1)π

20

β΄ 16 = β4 + 4π

β΄ 4π = 16 + 4 = 20 β΄ π =

=5

4

β΄ π = β4 + π = β4 + 5 = 1

β΄π = π+π = 1+5=6

β΄ π = 6 + 5 = 11

Then the three arithmetic means are: 1, 6, and 11

4. Find the sum of the arithmetic series for which a1 = 7, n = 31, and an= 127.

Solution:

π

31

The sum of the arithmetic series= (π1 + ππ ) = Γ (7 + 127) = ππππ

2

2

5. Find the next three terms of the geometric sequence 81, 27, 9β¦

Solution: the next three terms are: π, π,

π

π

6. Find the eighth term of the geometric sequence for which a1 = 5 and r = -2.

Solution: ππ = π1 Γ π (πβ1) = 5 Γ (β2)(8β1) = 5 Γ (β2)7 = βπππ

7. Find two geometric means between 7 and 189

Solution: Assume the two geometric means are x and y

Then the geometric sequence will be: 7, x, y, 189

π1 = 7, π4 = 189

ππ = π1 Γ π (πβ1)

β΄ π4 = π1 Γ π (4β1)

189

3

β΄ 189 = 7 Γ π 3

β΄ π3 =

= 27 β΄ π = β27 = 3

7

β΄ π₯ = π1 Γ π = 7 Γ 3 = 21

β΄ π¦ = π₯ Γ π = 21 Γ 3 = 63

Then the two geometric means are: 21 and 63

8. Find the sum of the geometric series for which a1= 125, r = β
, and n = 4

Solution:

The sum of the geometric series=

π1 (1βπ π )

1βπ

2 4

=

125Γ(1β(5) )

2

1β5

=...