njc math apgp lecture notes teachers edition

National Junior College Mathematics Department 2010

2010 / SH1 / H2 Math / APGP (Teacher’s Edition) of 18

National Junior College

2010 H2 Mathematics (Senior High 1)

Arithmetic and Geometric Series (Lecture Notes)

Topic 3: AP/ GP

Objectives:

At the end of this topic, students should be able to

� recognise if a sequence is an arithmetic progression or a geometric progression; � use the formula for the nth term of an arithmetic progression or geometric

progression; � use the formula for the sum to n terms of an arithmetic series, as well as a geometric

series;

� understand that 0 as when 1nr n r→ → ∞ < , and use it to deduce the sum to

infinity of a geometric series, and the condition for the sum to infinity to exist; � determine the finite sum of a series made up of arithmetic and geometric series; � solve practical problems involving arithmetic and geometric series.

§1 Introduction

At a job interview, you are asked to choose one the following two pay schemes: Scheme 1: A starting pay of $1500 per month and a monthly increment of $160 per year. Scheme 2: A starting pay of $1500 per month and a yearly increment of 10% of the

previous year’s salary. If you intend to stay in the company for 3 years, which pay scheme will you choose? Will you still choose the same pay scheme if you intend to stay in the company for 5 years?

Annual Salary Year 1 Year 2 Year 3 Year 4 Year 5

Scheme 1 $18000 $18000 + 12($160)

$18000 + 12($320)

$18000 + 12($480)

$18000 + 12($640)

Scheme 2 $18000 1.1($18000) 1.12($18000) 1.13($18000) 1.14($18000)

3 years:

Scheme 1: total pay = $18000 + [$18000 + 12($160)] + [$18000 + 12($320)] = $59760 Scheme 2: total pay = $18000 + [1.1($18000)] + [1.12($18000)] = $59580 Thus Scheme 1 is better for 3 years. 5 years:

Scheme 1: total pay = $18000 + [$18000 + 12($160)] + [$18000 + 12($320)] + [$18000 + 12($480)] + [$18000 + 12($640)] = $109200

Scheme 2: total pay = $18000 + [1.1($18000)] + [1.12($18000)] + [1.13($18000)] + [1.14($18000)] = $109891.80 Thus Scheme 2 is better for 5 years.



Hence you should choose scheme 2 if you intend to stay in the company for a longer period of time. Note that in scheme 1, the amount for each year forms a sequence known as arithmetic

progression because the difference in pay between any 2 consecutive years is the same. In scheme 2, the sequence generated is known as geometric progression since the ratio of the salaries of any 2 consecutive years is the same. Note: Writing down the first few terms may be a good method of seeing whether a sequence is an arithmetic progression or a geometric progression. §2 Arithmetic Progression

Definition

An arithmetic progression (A.P.) is a sequence of numbers in which each term other than the first term is obtained from the preceding one by the addition of a constant number called the common difference. Therefore, if we let a be the first term and d be the common difference of the sequence, then the sequence is an A.P. of the form

a, a + d, a + 2d, a + 3d, …… Can you deduce the formula for the general nth term of an A.P.?

Consider the following sequences of numbers. Do they form an A.P.? If yes, give the values

of a, d, and the general term nu .

Sequences First

Term, a

Common

Difference, d

General Term, nu A.P.?

1) 2, 4, 6, 8, 16, …. 2 - - No

2) 1,1 2,1 2 2,1 3 2,...+ + + 1 2 1 ( 1) 2n+ − Yes

3) 30,....,6,2,2,6 −−− 6 －4 6 ( 1)( 4)n+ − − Yes

4) ,...73,52,3 +++ πππ π +3 π + 2 3π + + (n – 1)(π + 2) Yes

Example 2.1

Find the number of terms in the following arithmetic progression: 2, 7, 12, 17, …, 77.

Solution:

First term, a = 2 Common Difference, d = 5 Let n be the number of terms.

Then, 77 2 ( 1)5 3 5 16n n n= + − = − + ⇒ =

( 1)nu a n d= + −



In an A.P., the difference between ANY 2 consecutive terms always gives the same

constant. Thus, to prove that a sequence is an arithmetic progression, we need to show that

Example 2.2

A sequence is given as follows: 8, 13, 18, …, 3 + 5n, … where n +∈Z . Show that this

sequence is an A.P.

Solution:

1

1

3 5

3 5( 1) 2 5

(3 5 ) ( 2 5 ) 5

n

n

n n

u n

u n n

u u n n

−

−

= +

= + − = − +

∴ − = + − − + =

which is a constant. Hence, the sequence is an A.P. Common Mistake:

2 1

3 2

13 8 5

18 13 5

u u

u u

− = − =

− = − =

thus sequence is an A.P.

Note: Do NOT merely use any two specific terms to prove a sequence is an A.P., use

general terms instead.

Example 2.3

The third term of an arithmetic progression is 10 and the seventh term is 34. Find the first term and the common difference. Hence find the twelfth term. Solution:

12

2 10 (1)

6 34 (2)

(2) (1) : 4 24 6

2

( 2) 11(6) 64

a d

a d

d d

a

u

+ = −

+ = −

− = ⇒ =

∴ = −

= − + =

1 constant, where denotes the general th term of the sequence.n n n

u u u n−

− =

This is not a substantial proof.



Example 2.4

Three consecutive terms of an arithmetic progression have sum 21 and product 315. Find the three numbers of the arithmetic progression. Solution:

Let the 3 terms be a – d, a, a + d.

2

2

( ) ( ) 21 7

(7 )(7)(7 ) 315

(7 )(7 ) 45

49 45

4 2

a d a a d a

d d

d d

d

d d

− + + + = ⇒ =

− + =

− + =

− =

= ⇒ = ±

Hence the three numbers are 5, 7, 9 or 9, 7, 5.

§3 Arithmetic Series

Let nS denote the sum of the first n terms of the arithmetic progression. Then

( ) ( ) ( ) ( ) [ ]1

2 3 ... 1 ( 1)n

n

i

S a a d a d a d a n d a i d=

= + + + + + + + + + − = + − ∑

where a is the first term and d is the common difference. nS is called an arithmetic series.

( )2 1 ( )2 2

n

n nS a n d a l= + − = +

where l is the last term, i.e. ( )dnal 1−+= .

Proof:

+( ) ( 2 ) ... ( ( 1) ) (1)

( ( 1) ) ( ( 2) ) ( ( 3) ) ... (2)

n

n

S a a d a d a n d

S a n d a n d a n d a

= + + + + + + − −

= + − + + − + + − + + −

Taking (1) + (2), we have

2 (2 ( 1) )

(2 ( 1) )2

( ( 1) )2

( )2

n

n

S n a n d

nS a n d

na a n d

na l

= + −

= + −

= + + −

= +



Example 3.1

Evaluate the following series: (i)

( ) [ ] [ ] [ ] [ ]

( )

( )

2

2 1 2(2) 1 2(3) 1 2(4) 1 ... 2( ) 1

5 7 9 ... (2 1)

2 15 2 1

2

12 6

2

n

r

r n

n

nn

nn

=

+ = + + + + + + + +

= + + + + +

− += + +

−= +

∑

(ii)

( ) ( )

( )

2

1

1 22

5 12

n

r n

nr n n n n n

nn

= +

+ = + + + +

= +

∑

Note that in both series above, the rth term forms an arithmetic progression. Thus we can easily apply the formula for finding the sum of an arithmetic series. Alternatively, you may apply the rules of summation (which you have learnt earlier) to evaluate the sum.

Example 3.2

The sum of the first nine terms of an arithmetic progression is 75 and the twenty-fifth term is also 75. Find the common difference and the sum of the first hundred terms.

Solution:

Let a be the first term and d be the common difference.

( )

100

92 8 75

2

254 (1)

3

24 75 (2)

200 10(2) (1) : 20 , 5

3 3

100 102( 5) 99 16000

2 3

a d

a d

a d

d d a

S

+ =

+ = −

+ = −

− = ⇒ = = −

= − + =

3.1 Finding “General Term Formula” from “Sum Formula”

The “sum formula” for an arithmetic series can be used to find a formula for its general term, nu .

Sn = 1 2 3 1.... n nu u u u u−

+ + + + +

Sn – 1 = 1 2 3 1.... nu u u u−

+ + + +

Therefore, Sn – Sn – 1 = nu



Example 3.3

The sum of the first n terms of a series is given by nnSn 32−= . Show that the terms of the

series are in an arithmetic progression. Hence, find the first term, a and the common difference, d. Solution:

2

2 2

1

2 2

1

1

1

3

( 1) 3( 1) 5 4

3 ( 5 4) 2 4

2( 1) 4 2 6

(2 4) (2 6) 2

n

n

n n n

n

n n

S n n

S n n n n

u S S n n n n n

u n n

u u n n

−

−

−

−

= −

= − − − = − +

∴ = − = − − − + = −

= − − = −

∴ − = − − − =

Hence, the terms are in A.P.

1 1 3 2a S= = − = − ; d = 2.

§4 Geometric Progression

Definition

A geometric progression (G.P.) is a sequence of numbers in which each term other than the first term is obtained from the preceding one by the multiplication of a non-zero constant

number called the common ratio. Therefore, if we let a be the first term and r be the common ratio of the sequence, then the sequence is a G. P. of the form

a, ar, ar2, ar

3, …… Can you deduce the formula for the general nth term of a G.P.?

Give the values of a, r, and the general term nu for the following geometric progressions:

Sequences First Term, a Common Ratio, r General Term, nu

1) 2, 6, 18, 54, …. 2 3 12(3 )n

nu−

=

2) 1 1 3 9

, , , ,....3 2 4 8

− − 1

3−

3

2−

11 3

3 2

n

nu

−

= − −

1n

nu ar−

=



Example 4.1

Find the number of terms in the following geometric progression: 1 1 1

8, 4, 2,1, , ,....,2 4 2048

.

Solution:

Let n be the number of terms.

First term, a = 8 Common ratio, r = 1

2

Then, 1

1

1 18

2048 2

1 1

16384 2

1 1ln ( 1) ln

16384 2

1 14 15

n

n

n

n n

−

−

=

=

= −

− = ⇒ =

In a geometric progression, the ratio of any 2 consecutive terms always gives the same constant. Thus, to prove that a sequence is a geometric progression, we need to show that

1

non-zero constantn

n

u

u−

=

Note: Use general terms to prove that a sequence is a G.P. and not specific terms.

Example 4.2

The nth term of a sequence is 13(2) ,n

nT n+ +

= ∈Z . Show that the sequence is a geometric

progression. Solution:

1

1

1

1

3(2)

3(2)

3(2)2

3(2)

n

n

n

n

n

n

n

n

T

T

T

T

+

−

+

−

=

=

= =

which is a non-zero constant. Hence, the sequence is a G.P.



Example 4.3

The first 3 consecutive terms of a geometric progression are 34,42,3 −−− kkk . Find the

value of k, the first term and the common ratio. Solution:

2

2 2

2 4 4 3

3 2 4

( 3)(4 3) (2 4)

4 3 12 9 4 16 16

7

k k

k k

k k k

k k k k k

k

− −=

− −

− − = −

− − + = − +

=

First term 7 3 4

2(7) 4 5Common Ratio

7 3 2

= − =

−= =

−

Example 4.4

Three consecutive terms of a geometric progression have a sum of 24.5 and a product of 343. Find the three numbers.

Solution:

3

2

Let the three terms be , , .

24.5 (1)

( )( ) 343

Since is non-zero,

343 7

Subst into (1)

7 497 7

2

2 5 2 0

(2 1)( 2) 0

1, 2.

2

aa ar

r

aa ar

r

aa ar

r

r

a a

rr

r r

r r

r

+ + = − −

=

= ⇒ =

∴ + + =

⇒ − + =

⇒ − − =

⇒ =

Hence, the three numbers are 14, 7, 7

2 or

7,7,14

2.



§5 Geometric Series

Let nS denote the sum of the first n terms of the geometric progression. Then

2 3 1 1

1

...n

n i

n

i

S a ar ar ar ar ar− −

=

= + + + + + =∑ .

where a is the first term and r is the common ratio. nS is called a geometric series.

nS =

Proof :

2 3 1

For 1,

... (1)n

n

r

S a ar ar ar ar −

≠

= + + + + + −

Multiply throughout by r,

2 3 ... (2)n

nrS ar ar ar ar= + + + + −

Taking (1) – (2):

(1 ) (1 )

1

1

n

n n

n

n

n

n

S rS a ar

S r a r

rS a

r

− = −

− = −

−=

−

For r = 1,

nS = a + a +… + a = an.

Example 5.1

The third term of a geometric progression is 36 and the sixth term is 121.5. Find the first term, common ratio and the sum of the first eight terms.

Solution:

2

3

5

6

36 (1)

121.5 (2)

u ar

u ar

= = −

= = −

Taking (2) ÷ (1):

Thus, 8

8

16(1 1.5 )788.125

1 1.5S

−= =

−.

n times

( )1, 1

1

, 1

na rr

r

an r

−≠

−

=

3

2

3.375

1.5

3616

1.5

r

r

a

=

=

= =



Example 5.2

The sum of the first n terms of a series is given by the expression 1

1

3

26

−

+

−n

n

. By finding an

expression for the nth term of the series, show that this is a geometric series. State the values of the first term and the common ratio. Solution:

1

1

1 2

1 1

1 1 2 1 2 2 2 1

1

1 2

26

3

26

3

2 2 2 2 2 2 1 2 26 6 1

3 3 3 3 3 3 3 3 3

2

3

n

n n

n

n n

n n n n n n n

n n n n n n n n n n

n

n n

S

S

u S S

u

+

−

− −

+ +

− − − − − − − −

−

− −

= −

= −

= − = − − − = − + = − + = =

=

21

1 1 1

12

2

1 0

1

22 3 23

2 3 2 3

3

Hence, it is a G. P.

2First term 6 2

3

2Common Ratio .

3

n

n nnn

n n n

n

n

n

n

u

u

S

T

T

−−

− − −

−

−

−

∴ = = × =

= = − =

= =

§6 Infinite Geometric Series

Consider the following sequences: 1, 2, 4, 8, 16, … a G.P. with first term 1 and common ratio 2.

1 1 1 1

1, , , , ,...2 4 8 16

a G.P. with first term 1 and common ratio 1

2.

1, －2, 4, －8, 16, … a G.P. with first term 1 and common ratio –2.

Let us find the sum of the first n terms for each sequence and denote the sum as nS .

(You may use your GC to generate the series)

Series nS 10S 50S 100S 200S

1 + 2 + 4 + 8 + 16 + … 2 1n− 1023 151.13 10× 301.26 10× 601.61 10×

1 1 1 11 ...

2 4 8 16+ + + + +

12 1

2n

−

1.998 2 2 2

1 + (－2) + 4 + (－8) +

16 + … ( )1 ( 2)

3

n− −

－341 143.75 10− × 294.23 10− ×

595.35 10− ×



For each of the series, what can you say about the value of nS as n → ∞ ?

Both the first and third series get larger (either positively or negatively) as n gets larger. But the second series gives the same value of 2 as n increases. Thus we say that the sum to infinity of the second series is finite and equals to 2. In other words, the series is convergent and it converges to the limit 2,

i.e. nS = 1

2 12n

−

→2 when n → ∞ (because

10 when

2nn→ → ∞ ).

When will the sum to infinity of a geometric series be finite? ,

Observe that for 1 1r− < < , 0 when .nr n→ → ∞

Hence, 1

when 1 1

n

n

r aS a n

r r

−= → → ∞

− − , and we say that the sum to infinity, denoted by

∞S , of the geometric series exists and

The geometric series is said to converge to the limit ∞

S .

Example 6.1

Evaluate the sum ...1000000

52

10000

52

100

52+++ . Hence, or otherwise, express the recurring

decimal of 3.525252…. as a rational number. Solution:

52 52 52 1 1 1... 52 ...

100 10000 1000000 100 10000 1000000

15210052

1 991

100

52 52 52 523.525252... 3 ... 3 .

100 10000 1000000 99

+ + + = + + +

= = −

= + + + + =

1

aS

r∞

=−

1 1, where denotes the common ratior r− < <

GP:

First term = 1

100

Common ratio

=1

100



Example 6.2

The sum to infinity of a geometric series is 162. The sum of the first 3 terms is 114. Find the least value of n for which the sum to n terms differs from the sum to infinity by less than 0.02. Solution:

3

3

3

3

162 162 (1)1

(1 )114 114 (2)

1

Substitute (1) into (2)

162(1 ) 114

114 8 21

162 27 3

Subst into (1)

2162(1 ) 54

3

aS

r

a rS

r

r

r r

a

∞= ⇒ = −

−

−= ⇒ = −

−

− =

⇒ = − = ⇒ =

= − =

0.02

2162 162 1 0.02

3

2162 1 1 0.02

3

2 0.02 2( ( ) 0 for all

3 162 3

modulus sign is removed)

2 0.02lg lg

3 162

0.02lg

162 22.22

lg3

n

n

n

n

n

S S

n

n

n

∞

+

− <

− − <

− − <

< > ∈

⇒

<

> =

∵ �

Therefore, the least n = 23.

Example 6.3 (Miscellaneous)

The nth term of a series is 12 5 2nn

−+ + . Find the sum of the first N terms.

Solution:

( ) ( )

( )( ) ( )

1 1 1

1 1 1 1 1

52 5 2 2 5 2 1 2 2 1 2

2

2 1 215 5 4 2 1 9 5

2 1 2 2 2

N N N N Nn n n

n n n n n

N

N

Nn n N N

N NN N

− − −

= = = = =

+ + = + + = + + +

−= + + + = − + +

−

∑ ∑ ∑ ∑ ∑

OR

Notice that 12n

nU

−= ( thn term of a G.P. with a = 1 & r = 2) and 5 2

nU n= + ( thn term of

an A.P. with a = 7)

2 1 (7 5 2) 2 1 (9 5 )

2 1 2 2

NN

N

N NS N N

−∴ = + + + = − + +

−



Example 6.4 (Miscellaneous)

The fifth, tenth and twentieth terms of a convergent geometric progression, G, are the first three consecutive terms of an arithmetic progression, A. (i) Determine the common ratio of G. Given that the first term of G is 2, (ii) evaluate the sum to infinity for G. (iii) find the sum of the first 10 odd-numbered terms in A.

[NJC/2008/H2Promo/Q8] Solution:

(i) 9 4 19 9 ar ar ar ar− = −

( ) ( )4 5 9 10

55

10

5

5

10 5

1 1

1

1

1

1

1 0

0.908

ar r ar r

rr

r

rr

r r

r

⇒ − = −

−⇒ =

−

⇒ =+

⇒ + − =

⇒ =

(ii) 2

21.8 1 0.908

S∞

= ≈−

(iii) Sum of the first 10 odd-numbered terms

( ) ( )( )4 9 4102 2 9 2 2 2

2

33.2

r r r = + −

= −



National Junior College

2010 H2 Mathematics (Senior High 1)

Arithmetic and Geometric Series (Tutorial)

Basic Mastery Questions

1. If the rth term of an arithmetic progression is 4r – 7, find the sum of the first 2n terms. 2. The first term of an arithmetic progression is 2. The sum of the first 8 terms is 58 and the

sum of the whole series is 325. Calculate (i) the common difference, (ii) the number of terms, (iii)the last term.

3. The first three terms, all positive, of a certain geometric progression are x – 4, x and 5x –

12. Find (i) the value of x,

(ii) the value of the fourth term.

4. The second term of a geometric progression is –18 and its sum to infinity is 81

2. Find the

value of the common ratio and the first term of the geometric progression.

Tutorial Questions

1. The rth term of a series is given by 21 3r− .

(i) Show that the series is an arithmetic series. (ii) Find the sum of the first n terms of the series. (iii)Find the nth term given that the sum of the first n terms is zero.

2. The sum of the first n terms of an arithmetic progression is given by 2qnpnSn += , and

that 63 =S and 115 =S .

(i) Find the values of p and q. (ii) Hence, or otherwise, find an expression for the nth term and the value of the common difference.

3. The sum of the first 20 terms of an arithmetic progression is 50 and the sum of the next

20 terms is –50. Find the sum of the first 100 terms of this progression. 4. An arithmetic progression has first term, a and common difference, 10. The sum of the

first n terms is 10000. Express a in terms of n, and show that the nth term is

( )10000

5 1nn

+ − . Given that the nth term is less than 500, show that

020001012<+− nn , and hence find the largest possible value of n.



5. A man earns $30000 for the first year and he receives a fixed annual increment of $100 for the second and each subsequent year. He deposits 20% of his yearly salary in a safe deposit box with a bank as part of his liquid assets. Calculate the amount of money in his safe deposit box after 15 years? (Note that any deposit in the safe deposit box does not yield any interest.)

6. The sum of the first n terms of a series is given by 1

25

5

n

n−− .

(i) Show that the series is a geometric series. (ii) Explain why this series converges and find its sum to infinity.

[DHS/2008/H2Prelim/P1/Q1]

7. An infinite geometric series has first term 1a − and second term 2 3 2a a− + . Given that all the terms of the series are positive, find the set of values of a for which the series converges.

Given that 11

4a = , find the least value of n for which the sum of the first n terms of

the series exceeds 6.999. [MJC/2008/H2Prelim/P2/Q2]

8. It is given that a, b, c are the first three terms of a geometric progression. It is also given

that a, c, b are the first three terms of an arithmetic progression.

(i) Show that 2 and 2

a bb ac c

+= = .

(ii) Hence show that

2

2 1 0b b

a a

− − =

.

(iii)Given that the sum to infinity of the geometric progression is S, find S in terms of a.

9. (a) An infinite geometric series has first term a and common ratio r. The sum of the first fourteen terms of the series is 127 times the sum to infinity of the remaining terms of the series. Find the two possible values of r in exact form.

(b) From a ribbon, pieces of decreasing lengths are cut. The lengths of the pieces cut follow an arithmetic progression with the 6th piece and the 26th piece cut being of

lengths 19 cm and 15 cm respectively.

(i) Find the length of the first piece cut and the common difference of the arithmetic progression.

(ii) Assuming that the ribbon is sufficiently long, find the number of such pieces that can be cut from the ribbon and also the least possible length of the ribbon.

[RI(JC)/2009/H2Promo/Q3]



10. A flag pole is being driven into the ground by a mechanical hammer. The distance it is driven by the first blow is 8 cm. Subsequently, the distance it is driven by each blow is

9

10 of the distance it was driven by the previous blow.

(i) The pole is to be driven a total of at least 70 cm into the ground. Find the smallest number of blows needed.

(ii) Explain why the pole can never be driven a total distance of more than 80 cm into the ground.

11. Benny, a self-employed businessman has a sum of money and he decides to make a

yearly contribution to his Central Provident Fund (CPF) Special Account as part of his investment plans. Interest is added to his Special Account at the end of each year at a rate of 4 % of the amount in the account at the start of that year.

Benny contributes $X to his Special Account at the beginning of one year and then deposits a further of $X to the same account at the start of each subsequent year. He also decides that he will not draw any money including the interest out of the account. He wishes to retire once he has saved more than $30X in his Special Account.

(i) Calculate how much money he will receive at the end of ten years? (ii) Find the least number of years he needs to continue contributing to his CPF Special Account before he can retire?

Challenging Questions

1. The positive integers, starting at 1, are grouped into sets containing 1, 2, 4, 8, … integers, as indicated below, so that the number of integers in each set after the first is twice the number of integers in the previous set.

{1}, {2, 3}, {4, 5, 6, 7} {8, 9, 10, 11, 12, 13, 14, 15}, …..

(i) Write down expressions, in terms of r, for

(a) the number of integers in the rth set, (b) the first integer in the rth set, (c) the last integer in the rth set.

(ii) Given that the integer 1000000 occurs in the rth set, find the value of r. (iii) The sum of all the integers in the 20th set is denoted by S, and the sum of all

the integers in all of the first 20 sets is denoted by T. Show that S may be

expressed as ( )18 192 3 2 1× − , and evaluate T

S, correct to 4 decimal places.



2.

‘Snowflake curves’ are generated by the following constructions:

Step 1: Start with an equilateral triangle, 0C .

Step 2: On the middle third of each side, draw an equilateral triangle and

erase the base of each of the new triangles (shown in 1C ).

Step 3: Step 2 is repeated to get figures 2C , 3C ,….

Let n

U denote the number of sides and n

P denote the perimeter of figure n

C .

Given that 0 3P = cm,

(i) find 0 1 2 3, , , and U U U U . Deduce n

U in terms of n, where n = 0, 1, 2, 3, …

(ii) find the least value of n such that n

C has more than 2010 sides.

(iii) find a relation between n

P and 0P , and hence, discuss the behaviour of

n

P as n tends to infinity.

(iv) find 0

k

nn

P=

∑ , where k+

∈Z . Explain if 0

nn

P∞

=

∑ converges?

Show that the area, n

A , enclosed by n

C , is 0

1 48 3

5 9

n

A

−

.

Discuss the behaviour of n

A as n tends to infinity.

Assignment Questions

1. The 10th term of an arithmetic progression is 20 and the sum of the first 20 terms is 500.

Find the sum of the first 50 odd-numbered terms.

2. The seventh, third and first term of an arithmetic progression with non-zero common difference are the first three terms of a geometric progression respectively. Prove that the geometric progression is convergent.

The seventh term of the arithmetic progression is 3. Find the smallest value of n such that the sum of the first n terms in the arithmetic progression exceeds the sum of the first n terms in the geometric progression by at least 100.

[AJC/2008/H2Prelim/P2/Q2]

0C 3C 2C 1C



3. Each time a ball falls vertically on to a horizontal floor, it rebounds to three-quarters of the height from which it fell. It is initially dropped from a point 4 metres above the floor. Find, and simplify, an expression for the total distance the ball travels until it is about to touch the floor for the (n + 1)th time. Hence, find the number of times the ball has bounced when it has travelled 24 metres and also the total distance it travels before coming to rest. (The dimensions of the ball are to be ignored)

Numerical Answers to Arithmetic and Geometric Series Tutorial Basic Mastery Questions

1. 28 10n n− 2. 3

(i) , (ii) 20, (iii)30.52

3. (i) 6, (ii) 54 4. 1

;543

−

Tutorial Questions

1. (39 3 )

(ii) , (iii) 132

n n− 2. (i) p = 1.7, q = 0.1, (ii)0.2n + 1.6; 0.2 3. －750

4. 10000

5( 1)nn

− − ; 73 5. $92100 6. (ii) 5 7. { }: 2 3a a∈ < <� ; 31

8. (iii) 2

3

aS = 9. (a)

1

2r = ± , (b)(i) 20cm, −0.2cm, (ii) 100, 1010cm

10. (i) 20 11. (i) $12.49X, (ii) 20

Challenging Questions

1. (i) 1 12 ;2 ;2 1r r r− −− , (ii) 20, (iii) 1.3333

2. (ii) 33 (iii) 0

4

3

n

nP P

=

(iv)

1

0

49 1

3

kk

nn

P

+

=

= −∑

njc math apgp lecture notes teachers edition

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