Functions II (solucionario)Material AICLE 3º E.S.O: Material Functions II (Solucionario) 5 3) a)...

Material AICLE Secundaria 2º E.S.O. : ANCIENT GREECE 3Material AICLE 3º E.S.O: Functions II (Solucionario)

SOLUTIONS

A sequence of ideas to start

2)

The situations given before are related to a concept known as sequence.

A sequence is an ordered list of things (objects, events…).

Like a set, it contains members (they can be called elements or terms), and

the number of terms (possibly infinite) is called the length of the sequence.

Unlike a set, order matters, and the exact same

elements can appear several times at different positions in the sequence.

4) It is easy:

a) There is more than one possible correct option. Really every option would be

right! b) Although there can be more options, the expected answer is:

IV) 5, 15 ,25 (or 45),…

v) 2,4,8,…

5) a) We need to choose the first term. If we “start” by 5, then we have 5,10,20,40,80. If the

first term is 3: 3,6,12,24,48,… and so on. b) 1,3,5,7,9 c) 1.4, 1.43, 1.46, 1.49, 1.52

6) a) The table is:

Sentences (to complete) Sequences (to write)

Every term is double the previous one and the sequence starts by 10.

The second term is 20 and every term is the previous one plus 5.

The second term is the square root of sixteen and every term is three times the previous one.

Every term is the sum of the two previous terms and the sequence starts 3,4.

10, 20, 40, 80,…

15, 20, 25, 30, 35,…

4/3,4,12,36,…

3, 4, 7, 11, 18,…

Material AICLE Secundaria 2º E.S.O. : ANCIENT GREECE4 Material AICLE 3º E.S.O: Functions II (Solucionario)

b) The right answers are: by adding 3 by multiplying by 3

by multiplying by -‐1, or by changing the sign. by subtracting 4

8) a) an=6·∙2n-‐1

b) an= 3+(n-‐1)·∙4 c) an= (-‐1)

n+1

d) an=an-‐1+2n-‐1 (and a1=1)

“Increasing” what you know

1) The true options are:

On sequence a) every term is greater than the previous one. On sequence b) every term is greater than the next one.

2) The text is:

A sequence is increasing if every term is greater the previous one and a sequence is decreasing if every term is less than the previous one.

4) a) 4, 3.125, 3.037…, 3.0156…, 3.008, 3.0046…

b) It decreases. c) The terms get close to three!

5) The limit of bn is 2 and the limit of fn is 0. There is no possible limit for the other sequences.

In progression!

2) The solution is:

5, 8, 11, 14, … 5, 11, 17, 23, … 5, 3, 1, -‐1, …

-‐2 6 3


3) a) The sentences are:

Some sequences are special because they have a singular structure. The difference of any two successive terms of the sequence is a constant. They can be defined by the first term and the constant value called common difference.

These sequences always increase (or decrease). These sequences are called arithmetic progressions.

b) The solution is:

C) The terms are: • 5, 14, 23, 32, … • -‐3, -‐1, 1, 3, …

• 0, 0.3, 0.6, 0.9, … • 8, 3, -‐2, -‐7

5) The answer is: • 2,5,8, … • 9 steps

• d= 3 • 3x9=27

• 3x9 = 27 • a10= 2+27=29 • an= a1 + (n-‐1)·∙d

6) a) 1420 b) 518

c) -‐392.75 d) -‐6.5

The right arithmetic progressions are: 2,4,6,8,10, …

4, 4.5, 5, 5.5, … 9, 4, -‐1, -‐6, …

The common difference is: 2

0.5 -‐5


Another kind of progression!

2) The complete chart is:

Sentences (to complete) Sequences (to write)

Every month you recycle three times what you recycled last month. You started whit 0.5kg of garbage.

0.5, 1.5, 4.5, 13.5, …

Your study time has doubled every week since you studied only one hour a week.

1, 2, 4, 8, …

Your business produces every year five times the profits you obtained last year. At the beginning you earned 50€.

50€, 250€, 1250€, 6250€, …

3) a)

4) The complete text is:

A geometric progression is a sequence where every term is generated by multiplying the previous one by a fixed number called the common ratio of the progression.

Example: if a1 = 3 and the common ratio is r=2, the first terms are:

3, 3·∙2=6, 6·∙2=12, 12·∙2=24…

On this new kind of sequence

every term is calculated by

multiplying the previous number by certain value.

The certain value

is the same every time, is constant.

The terms you obtain are

decimal numbers sometimes.


5) The fifth term is:

Geometric progression: Fifth term:

2, 10, 50, … 1250

25, 75, … 2025

-‐4, 12, -‐36, … -‐324

4, 2, … 0.25

6)

2,6,8,… Four “steps”. You must multiply by r=3.

You multiply by 34=81. a5= 162. an= a1 ·∙ r

n-‐1

7) The results are:

a) 13671875

b) 2324522934 c) -‐98415 d) 0.00390625

If a little boy could

2) a)

b)

c)

d)

S = 1+2+3+……………..+198+199+200

1 + 2 + 3 +…………………..+198+199+200

200+199+198+……………………+ 3 + 2 + 1

1 + 2 + 3 +…………………..+198+199+200

200+199+198+……………………+ 3 + 2 + 1

201+201+201+……………………+201+201+201

+

2S = 201·∙200 = (1+200) ·∙ 200


e)

3) Previous investigation: if p+q=s+t…

ap+aq = a1+(p-‐1)d + a1+(q-‐1)d = 2a1+ (p+q-‐2)d =2a1+ (s+t-‐2)d =

a1+(s-‐1)d + a1+(t-‐1)d = as+at

That’s why if you sum Sn=a1+ a2+ a3+….. an-‐2+ an-‐1+ an and Sn= an+ an-‐1+ an-‐2+…… a3+ a2+ a1 you

have n times the sum a1+an. So 2Sn=(a1+an)n and finally:

4) The text is:

The formula to calculate the sum of the first n terms of in arithmetic progression (Sn) is the result of doing this:

• Adding the first and the last term you want to sum • Multiplying by the number of terms you are adding. • Dividing everything by two.

Example: if you want to add the 20 first terms of the arithmetic progression with first term a1=3 and common difference d=5…

; a20=3+19·∙5=98; =1010

5) The solution is:

a1 d an Sn

2 6 a10=58 290

-‐15 86 a157=13401 1050801

-‐4782 98 a50=20 -‐119050


A higher progression

2) The solution is:

a1 d a10 S10

2 7 80707214 94158416

-‐3 10 -‐0.000000003 -‐3333333333

5 1/3 2.5402631… 0.000381…

Functions II (solucionario)Material AICLE 3º E.S.O: Material Functions II (Solucionario) 5 3) a)...

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