Chapter 1 Text work review

10. Determine S₂₅.

     t₂ = 40, t₅ = 121     d = ?     t₁ = ?    t₂₅ = ?

      t₅ = t₂ + 3d  --> 121 = 40 + 3d --> 3d = 81 ---> d = 27

     t₂ = t₁ + d ---> t₁ = 13

     t₂₅ = 13 + 27(24) = 13 + 648 --> t₂₅ = 661

       S₂₅ = \( \frac{25}{2} \)(13 +  661) = \( \frac{25}{2} \)(674) = 8425

11. Find the first four terms:

   S₅ = 85   S₆ = 123   t₆ = ?   d = ?

    t₆ = S₆ -  S₅ = 123 - 85 = 38

   S₆ = \( \frac{6}{2} \)(t₁ + 38) = 123 ---> t₁ + 38 = 41 ---> t₁ = 3

    t₆ = 3 + 5d = 38 --> 5d = 35 --> d = 7

    so, the first four terms are 3, 10, 17, 24

12. Derive d = 5n².

time (s)	1	2	3	4
distance fallen in this second (m):	5	15	25	35
Total distance fallen:	5	20	45	80

We know that t₁ = 5, and t₂ = S₂ - S₁ = 20 - 5 = 15 = t₂

This gives us d = t₂ - t₁= 15 - 5 = 10 = d

and then, tn = t₁ + d(n - 1) = 5 + 10(n - 1) = 5 + 10n - 10 = 10n - 5 = t_n

Using S_n = \( \frac{n}{2} \)(t₁ + t_n) we get

          S_n = \( \frac{n}{2} \)(5 + 10n - 5) = \( \frac{n}{2} \)(10n) or

          S_n = 5n² = d

16. Distance from top to bottom...

First, let's be sure how many rings there are, according to diameter:

20, 19, 18, ..., 3

here, t₁ = 20, d = -1, and t_n = 3

3 = 20 - (n - 1) ---> n = 18, so there are 18 rings.

From the diagram, one can see that after the first ring, the distance from the bottom of one ring to the bottom of the next follows the pattern: 17, 16, 15, ..., 1, where there are 17 terms, with t₁ = 17, and t_n = 1, n = 17.

We want 20 + S₁₇ = 20 + \( \frac{17}{2} \)(17 + 1) = 20 + 153 = 173cm = distance from top to bottom.

1. If it is a geometric sequence, state the common ratio r, and the general term t_n.

a) 1, 2, 4, 8, ...

   r = 2, t_n = 2^{n - 1}

b) 2, 4, 6, 8, ... not a geometric sequence

c) 3, -9, 27, -87, ...

   r = -3, t_n = 3(-3)^{n - 1}

d) 1, 1, 2, 4, 8, ... not a geometric sequence

e) 10, 15, 22.5, 33.75, ...

   r = 1.5, t_n = 10\( \cdot \)1.5^{n - 1}

f) -1, -5, -25, -125, ...

   r = 5, t_n = -5^n-1

3. Determine the first four terms:

a) t₁ = 2, r = 3

   2, 6, 18, 54

b) t₁ = -3, r = -4

   -3, 12, -48, 192

c) t₁ = 4, r = -3

   4, -12, 36, -108

d) t₁ = 2, r = 0.5

   2, 1, 0.5, 0.25

5. Determine the formula for the general term:

a) r = 2, t₁ = 3

   t_n = 3\( \cdot \)2^{n - 1}

b) 192, -48, 12, -3, ...

   r = -\( \frac{1}{4} \), t₁ = 192, t_n = 192(\( \frac{1}{4} \)^{n - 1}

c) t₃ = 5, t₆ = 135

   t₆ = t₃r³, r³ = \( \frac{135}{5} \) = 27 = 3³ ---> r = 3

   t₃ = t₁\( \cdot \)3² ---> t₁ = \( \frac{5}{9} \)

   t_n = \( \frac{5}{9} \)\( \cdot \)3^{n - 1}

d) t₁ = 4, t₁₃ = 16,384

   t₁₃ = 4r¹² = 16,384 ---> r¹² = 4096 = 2¹² ---> r = 2

   t_n = 4\( \cdot \)2^{n - 1} = 2²\( \cdot \)2^{n - 1} = 2^{n + 1}  = t_n 

9. Fan out system:

t₁ = 1, r = 4

a) 1 + 4 + 16 + 64 + ...   (if person in charge is included)

    4 + 16 + 64 + ...   (if person in charge is not included)

b) Let's go with the person in charge not being included, and find out S₁₀.

In this case, t₁ = 4, r = 4:

S₁₀ = 4(4¹⁰ - 1)/(4 - 1) = 1,398,100 people contacted.

10. 

The ball will drop 6 times, that is, n = 6, where t₁ = 20, and r = 0.4.

The ball will rise 5 times (n = 5), with t₁ = 0.4(20) = 8, and r = 0.4.

To find the total vertical distance travelled, there are a few different ways we could look at this.

First, we could find the sum of the two series: down and up:

S_down6 + S_up5 = 20(0.4⁶ - 1)/(0.4 - 1) + 8(0.4⁵ - 1)/(0.4 - 1) = 33.2 + 13.2 = 46.4m

Or... since after the first drop, the distance up is the same as the distance down, we could look at ...

20 + 2S₅ where t₁ = 8, r = 0.4 --> distance = 20 + 2\( \cdot \)8(0.4⁵ - 1)/(0.4 - 1) = 20 + 2\( \cdot \)13.2 = 46.4m

11. Marathon training: t₁ = 25km, r = 1.1, n = 15, S₁₅ = ?

S₁₅ = 25(1.1¹⁵ - 1)/(1.1 - 1) = 794.3km

12. Koch snowflakes:

length: t_n = (\( \frac{1}{3} \))^{n - 1}

# segments: t_n = 3(4)^{n - 1}

perimeter: t_n = 3(\( \frac{4}{3} \))^{n - 1}

d) perimeter at stage 6 (n = 6)

t₆ = 3(\( \frac{4}{3} \))^{6 - 1} = \( \frac{1024}{81} \) = 12.64

13. Advertising company: t₁ = 1000, r = 1.4, want S₁₁.

n is a counter and when n = 2, it represents the next 10 days. We want 100 days, so n = 11:

S₁₁ = 1000(1.4¹¹ - 1)/(1.4 - 1) = 98739 people.

14. 10 beads: t = 24mm, r = 0.75, n = 10, S₁₀ = ?

S₁₀ = 24(0.75¹⁰ - 1)/(0.75 - 1) = 91mm

15. 200mg ampicillan ear drugs: t₁ = 200, r = 0.12

a) S₃ = 200(0.12³ - 1)(/(0.12 - 1) = 226.9mg

b) S₆ = 200(0.12⁶ - 1)/(0.12 - 1) = 227.3mg

16. t₁ = 3, r = 3, S_n = 9840, n = ?

9840 = 3(3ⁿ - 1)/(3 - 1) 

9840(\( \frac{2}{3} \)) = 3ⁿ - 1 = 6560

3ⁿ = 6561 = 3⁸

n = 8

17. t₃ = 24, t₄ = 36, S₁₀ = ?

t₄ = rt₃

r = \( \frac{36}{24} \) = \( \frac{3}{2} \)

t₃ = r²t₁

t₁ = 24(\( \frac{2}{3} \))² = 24\( \frac{4}{9} \) = \( \frac{32}{3} \)

S₁₀ = \( \frac{32}{3} \)(\( \frac{( \frac{3}{2})^9 \frac{3}{2} - 1 }{ \frac{3}{2} - 1 } \)) = \( ( \frac{32}{3} )( \frac{ \frac{59049}{1024} - 1 }{ \frac{1}{2} }) = (\frac{64}{3} )( \frac{59049}{1024} - \frac{1024}{1024}) \) = \( ( \frac{64}{3})( \frac{58025}{1024}) \) = \( \frac{58025}{48} \)

18. a + b + c = 35, abc = 1000, what are a, b, c?

b = ar, c = br = ar²

abc = 1000 

a(ar)(ar) = 10³

a³r³ = 10³

ar = 10

r = \( \frac{10}{a} \)

b = ar = a\( \frac{10}{a} \) = 10

a + b + c = 35  

a + 10 + ar² = 35

a + a(\( \frac{100}{a^2} \) = 25

a² - 25a + 100 = 0

(a - 20)(a - 5) = 0

a = 20 or a = 5

a + c = 25, c = 25 - a

c = 5 or c = 20

a = 20, b = 10, c =5   OR   a = 5, b = 10, c = 20

19. S₇ = 89, S₈ = 104, t₈ = ?

t₈ = S₈ - S₇ = 15

20. 

1.

4. Does 0.9999... = 1?

0.9999.... = 0.9 + 0.09 + 0.009 + ...

t₁ = \( \frac{9}{10} \) , r = \( \frac{1}{10} \)

S\( _\infty \) = \( \frac{ \frac{9}{10} }{1 - \frac{1}{10} } = \frac{ \frac{9}{10} }{ \frac{9}{10} } = 1 \)

10. S\( _\infty \) = 2t₁. r = ?

S\( _\infty \) = 2t₁ = \( \frac{t_1}{1 - r} \)

2(1 - r) = 1

2 - 2r = 1

2r = 1

r = 0.5

11. For what values of x will each series be convergent?

a) 5 + 5x + 5x² + 5x³ + ...

r = x, t₁ = 5

-1 < x < 1

b) 1 + \( \frac{x}{3} \) + \( \frac{x^2}{9} \) + \( \frac{x^3}{27} \) + ...

r = \( \frac{x}{3} \)

-1 < \( \frac{x}{3} \) < 1

-3 < x < 3

c) 2 + 4x + 8x² + 16x³ + ... 

r = 2x

-1 < 2x < 1

-\( \frac{1}{2} \) < x < \( \frac{1}{2} \)

12.