Maths ch16 key

p.1

Chp 16 Key Prepared by C.Y. So

e.g. 1 p.2

It is given that the first term is 6 and the third term is 2.

Let a = first term and d = common difference.

Thus, 6a and 22 da .

It follows that 22)6( d

4

82

622

22)6(

d

d

d

d

(a) Now, the general term dnanT )1()(

n

n

n

410

446

)4)(1()6(

(b) Using (a), )15(410)15( T

50

6010

)15(410

e.g. 2 p.3

It is given that the 175 T and 7720 T .


Thus,

)2( ...... 174

)1( ...... 7719

da

da.

(1)(2), 6015 d

4d

Sub 4d into (2), 17)4(4 a

1a

The 1st term is 1. The 2nd term = 1 + 4 = 5. The 3rd term = 5 + 4 = 9. The 4th term = 9+4 = 13.

p.2

15 23

d d d d

e.g. 3 p.3

It is given that the 501 T , 472 T and 443 T .


Thus, 50a and 5047 d

3

(a) Let 17)( nT . (b) Let 0)( nT .

12

336

31753

173350

17)3)(1()50(

17)1(

n

n

n

n

n

dna

3

217

3

53

353

03350

0)3)(1()50(

0)1(

n

n

n

n

n

dna

Thus, the 12th term is 17. Thus, the 17th term is positive and the 18th term is negative.

2

4850

)3(16)50(17

T

Thus, the smallest positive term is 2.

(Actually, there is a much quicker way. Think of it yourself.)

e.g. 4 p.4

The difference between 15 and 23

= 23 (15)

= 38

Let d = the common difference between two terms.

5.9

384

d

d

The first term added in The second term added in The third term added in

5.5

5.915

4

5.95.5

5.13

5.94

p.3

20 40

d d d d d d

5 27

d d d

x y

e.g. 5 p.5


= 40 20

= 20


3

13

3

10

206

d

d

d

The first term added in The second term added in The third term added in

3

123

3

1320

3

226

3

13

3

123

303

13

3

226

The fourth term added in The fifth term added in

3

133

3

1330

3

236

3

13

3

133

e.g. 6 p.5


= 27 5

= 22


3

17

3

22

223

d

d

d

3

112

3

175

x

3

219

3

17

3

112

y

p.4

e.g. 7 p.6

By formula: Recall that with a = first term, d = common difference and n = number of terms

Sum to n terms ])1(2[2

)( dnan

nS

For ...131074 ,

424

538

)3158(8

)]3)(116()4(2[2

)16()16(

S

Alternative: 49...131074

424

161363

162

13123

16)12...4321(3

16)48...12963(

e.g. 8 p.6

Let the three numbers be dy , y and dy .

As the sum of them equals 18, As the product of them equals 210,

6

183

18)()(

y

y

dyydy

1

1

3536

35)6)(6(

210)6)(6)(6(

2

2

d

d

d

dd

dd

Therefore, the three numbers are 5, 6 and 7.

e.g. 9 p.6

For the G.S., let a = the first term and r = the common ratio.

As the sixth term is 1, 15 ar .

As the third term is 8, 82 ar .

p.5

(a) Thus, we have 8

12

5

ar

ar

2

1

8

1)(

8

1

3

1

3

13

3

r

r

r

The common ratio is 2

1 .

(b) Sub 2

1r into 82 ar .

32

84

1

82

12

a

a

a

The first term is 32.

(c) The tenth term

16

12

1)1(

2

132)1(

2

1)32(

4

9

9

9

ar

e.g. 10 p.8


As the seventh term is 48, 486 ar .

As the fourth term is 6, 63 ar .

Thus, we have 6

483

6

ar

ar

p.6

2

8)(

8

3

13

13

3

r

r

r

Sub 2r into 486 ar .

4

364

48

4864

482

48)2(6

6

a

a

a

a

a

Thus, the first term is 4

3 and the common ratio is 2.

e.g. 11 p.8


It is given that the sum of the 1st term and the 3rd term is 40.

402 ara … (1)

It is given that the 2nd term is larger than the 1st term by 8.

8 aar … (2)

From (1), 402 ara

40)1( 2 ra

From (2), 8 aar

8)1( ra

Thus, we have 8

40

)1(

)1( 2

ra

ra.

5)1(

)1( 2

r

r

065

551

)1(51

2

2

2

rr

rr

rr

2r or 3r .

When 2r , 8a . In this case, the first term is 8 and the common ratio is 2.

When 3r , 4a . In this case, the first term is 4 and the common ratio is 3.

p.7

e.g. 12 p.9


After adding in five terms between 2 and 1458,

2 is still the 1st term and 1458 becomes the 7th term.

Thus, we have 14586 ar … (1)

and 2a … (2).

(1) (2), we have

3

729

2

1458

6

6

r

r

a

ar

The first number added in The second number added in The third number added in

6

32

18

36

54

318

The fourth number added in The fifth number added in

162

354

486

3162

e.g. 13 p.9


It is given that x, 6, 18, y are in a G.S.

3

186

r

r

2

63

x

x

54

54

318

y

y

y

p.8

e.g. 14 p.10

By formula: Recall that with a = first term, r = common difference and n = number of terms

Sum to n terms )1(1

)( nrr

anS

For ...1263 , 2r .

3069

)11024(3

)12(12

3

])2(1[)2(1

)3()10(

10

10

S

Alternative: 923...1263

3069

)11024(312

123

21

213

)21(

)21)(2...221(3

)2...221(3

23...232313

10

10

92

92

92

e.g. 15 p.10

For the G.S., 15, 30, 60, … , 3840,

let a = the first term and r = the common ratio.

(a) Thus, 15a .

2

3015

r

r

Let 38401 nar .

9

81

22

2562

15

3840

15

2)15(

38402)15(

81

1

1

1

n

n

n

n

n

n

The number of terms of the G.S. is 9.

p.9

(b) )3840(...)60()30()15(

7665

51115

)12(12

)15(

)21()2(1

)15(

9

9

e.g. 16 p.11

By formula: Recall that with a = first term and r = common difference

Sum to infinity r

aS

1

For ...139 , 39 r

3

1r

2

272

393

2 9

3

11

)9(

S

e.g. 17 p.11

By formula: Recall that with a = first term and r = common difference

Sum to infinity r

aS

1

4

4164

1

16

4

31

16

a

a

a

a

p.10

e.g. 18 p.12

(a)

6

1 36

i

i

29

7362

76

3

136

)654321(3

166

3

)6(6

3

)5(6

3

)4(6

3

)3(6

3

)2(6

3

)1(6

(b)

n

i

inS

1 36)(

36

12482

98

3

148

)8...321(3

186

3

16

36

36)8(

8

1

8

1

8

1

8

1

8

1

ii

ii

i

i

i

iS

Ex16.4 1. p.13


As the 9th term is 25, As the 5th term is 13,

258 da … (1) 134 da … (2)

(1) (2), 13254 d

124 d

3d

13)3(4 a , 1a .

The 60th term = (1) + (59)(3)

= 178

p.11

Ex16.4 2. p.13

(a) For the A.S., 1002, 1005, 1008, … , 1998

can be rewritten as (1000+2), (1000+5), (1000+8), … , (1000+998)



Therefore, there are 333 terms in the A.S. 1002, 1005, 1008, … , 1998 which are divisible by 3.

(Alternative: You may let 1002a , 3d , 1998)1()(T dnan and solve for n)

(b) For the A.S., 1000, 1004, 1008, … , 2000

can be rewritten as 1000, (1000+41), (1000+42), … , (1000++4250)


(c) For the A.S., 1008, 1020, 1032, … , 1992



Thus, the number of integers between 1000 and 2000 (including 1000 and 2000)

501

83251333

Ex16.4 3. p.13

(In solving this problem, the following technique has been employed : 2231 TTTT )

As a is the arithmetic mean between 8 and b, As 12 is the arithmetic mean between a and b,

it means 8, a, b forms an A.S.. it means a, 12, b forms an A.S..

aab 8 (Using the fact that 2231 TTTT )

1212 ba

82 ba … (1) 24 ba … (2)

(1) + (2), 2483 a

3

32a

3

210a

Sub 3

210a into (2), 24

3

210

b

3

113b

p.12

Ex16.4 4. p.13


As the sum of the second and third term is 20, As the sum of the fourth and fifth term is 320,

202 arar 32043 arar

20)1( rar ... (1) 320)1(3 rar … (2)

(2) (1), 20

320

)1(

)1(3

rar

rar

4

162

r

r

Sub 4r into (1), 20)5)(4( a

1a

The general term T(n)

1

1

1

4

)4)(1(

n

n

nar

Ex16.4 5. p.14

Let the three numbers be a, da and da 3 .

(a) As they form a G.S., ))(()3)(( dadadaa (Using the fact that 2231 TTTT )

2

222 23

dad

dadaada

02 dad 0)( dad Thus, 0d or da .

As the sum of the three numbers is 14,

1443

14)3()(

da

dadaa

When 0d ,

3

14

143

a

a

Thus, the three numbers are 3

14,

3

14,

3

14.

The common ratio is 1.

p.13

When ad ,

2

147

a

a

Thus, the three numbers are 2, 4, 8.

The common ratio is 2.

(b) The common difference of the A.S. is 0 or 2.

(c) When 0d , the three numbers are 3

14,

3

14,

3

14.

When 2d , the three numbers are 2, 4, 8.

Ex16.4 6. p.14

(In solving this problem, the following technique has been employed :

2231 TTTT for an A.S.

2231 TTTT for a G.S.)

As 1, x, y are in a G.S., As x, y, 15 are in an A.S.,

2

1

xy

xxy

152

15

yx

yyx

Combining the above two equations,

1520

152

15)(2

2

2

2

xx

xx

xx

2

5x or 3x

When 2

5x , When 3x ,

2

2

5

y

9

)3( 2

y

4

25

Ex16.4 7. p.14

(This question is cancelled)

p.14

Ex16.4 8. p.14

By observation: 2, 22, 204, 2006, 20008, 200010, …

= 2, (20+2), (200+4), (2000+6), (20000+8), (200000+10), …

= 2, (20+2), (200+4), (2000+6), (20000+8), (200000+10), …

The n th term is )1(2102 1 nn

The sum of the first n th terms

)]1(2102[...)10200000()820000()62000()4200()220(2 1 nn

)]1(2...6420[)102...200000200002000200202( 1 nn

)]1(...321[2)10...101010101(2 1432 nn

)1()110(9

22

)1(2

101

1012

nn

nn

n

n

Alternative:

(You are strongly recommended NOT to use the following approach. However, the way of handling

a sequence may give you insight of handling an unknown sequence.)

For the sequence 2, 22, 204, 2006, 20008, 200010, …

The differences of between two consecutive numbers are

20, 182, 1802, 18002, 180002, …

which can be rewritten as

21018 0 , 21018 1 , 21018 2 , 21018 3 , 21018 4 , …

Thus, for the sequence 2, 22, 204, 2006, 20008, 200010, …

2T1 ,

)21018(2T 02 ,

)21018()21018(2T 103 ,

)21018()21018()21018(2T 2104

)21018()21018()21018()21018(2T 32105

…

)21018(...)21018()21018()21018()21018(2T 23210 nn

p.15

Actually,

1

1

1

1

1

2321

2321

23210

102)1(2

21022

)110(22

110

110182

101

)101(182

101

)101()10...1010101(182

)10...1010101(182

)1018(...)1018()1018()1018()1018(2T

n

n

n

n

n

n

n

nn

n

n

n

n

n

n

n

n

Now,

n

iinS

1

T)(

)110(9

2)1(

101

1012

2

)1(2

)10...10101(2))1(...3210(2

102)1(2

102)1(2

]102)1(2[

12

1

1

1

1

1

1

1

1

n

n

n

n

i

in

i

n

i

in

i

n

i

i

nn

nn

n

i

i

i

Ex16.4 9. p.15

(a) The total distance moved just before the second rebound

m 254

310

4

31010

(b) The total distance moved just before the (k+1)th rebound

p.16

1

1

1

1

1

2

2

22

4

38070

4

3

4

38010

1

4

3

4

34

2010

4

14

3

4

3

2010

4

31

4

3

4

3

2010

4

3...

4

3

4

32010

4

320...

4

320

4

32010

4

310

4

310...

4

310

4

310

4

310

4

31010

k

k

k

k

k

k

k

kk

(c) As k tends to infinity, 1

4

3

k

tends to zero.

The total distance moved travelled by the ball before it comes to rest = 70 m

Ex16.4 10. p.16

(a) (i) 3

1%)1(1 rPQ

%)1(3

rP

3

1%)1(

3

2%)1(2

rrPQ

2

2

%)1(9

23

2%)1(

3

rP

rP

p.17

(ii) 3

1%)1(

3

2%)1(

3

2%)1(3

rrrPQ

3

23

%)1(27

4

3

2%)1(

3

rP

rP

(b) 1Q , 2Q , 3Q , … is a G.S.

The common ratio

%)1(3

21

%)1(9

23

3

1

%)1(9

2

%)1(3

1

%)1(9

2 2

1

2

r

r

r

rP

rP

Q

Q

(c) (i) PQ128

273

125.1%18

9%1

2

3%1

2

3%)1(

22

33%)1(

4

27

128

27%)1(

128

27%)1(

27

4128

27%)1(

27

4

3

2

9

63

27

333

3

3

3

r

r

r

r

r

r

r

PrP

Thus, r = 12.5

p.18

(ii) 10321 ... QQQQ

99

92

2

2

109

93

2

22

1032

%)1(3

2...%)1(

3

2%)1(

3

21%)1(

3

1

%)1(3

2

3

1...%)1(

3

2

3

1%)1(

3

2

3

1%)1(

3

1

...%)1(27

4%)1(

9

2%)1(

3

1

rrrrP

rPrPrPrP

QrPrPrP

)...1%)(1(3

1 932 yyyyrP (Take %)1(3

2ry )

y

yrP

1

1%)1(

3

1 10

With 10000P , 125.1%1 r , 125.13

2y

75.0

14155

29728.1415575.01

75.01125.110000

3

1

...10

10321

QQQQ

Ex16.4 11. p.17

(a) For the geometric series ...25088

3

2818

3

24

33

,

Let a = the first term and r = the common ratio.

3a . As 24

3

24

13

, 24

1

r .

163.2

1633634.2

24

11

24

11

3

1

)1(

)...1(

...

20

20

192

192

r

ra

rrra

ararara

(b) r

aS

1

p.19

)218(23

3

)2220(23

3

)25()25(

)25()24(3

25

)24(3

124

)24(3

24

11

3

24

11

3

Ex16.4 12. p.17

(a) 9, 99, 999, 9999, …

= )110( , )110( 2 , )110( 3 , )110( 4 , …

The n th term is )110( n

(b) Sum to n terms

9

10910

9

9

9

1010

110

1010

101

1010

)10...101010(

)110(...)110()110()110(

1

1

1

1

32

32

n

n

n

n

n

n

n

n

n

n

n

Ex16.4 13. p.17

(a) ...151173

can be considered as ...)116()112()18()14(

Sum to n terms

p.20

nn

nnn

nnn

nnn

nn

n

2

2

2

22

)1(22

)1(4

)...321(4

)14(...)144()134()124()14(

(b) When 820nS ,

08202

82022

2

nn

nn

20n or 5.20n (rejected)

Thus, n = 20.

(c) When 1500nS ,

015002

150022

2

nn

nn

637.27n or 137.27n

(rejected)

Thus, the smallest value of n is 28.

Ex16.4 14. p.18

Let the marks of the 1st, 2nd and the 3rd questions be 3y , y and 3y respectively.

It is given that the sum of the marks of the first 3 questions is 12.

4

123

12)3()3(

y

y

yyy

The marks of the questions are 1, 4, 7, 10, 13, …

The sum of the marks of the 2nd and 5th

= 4 + 13

= 17

Thus, the boy got 17 marks.

Ex16.4 15. p.18

(a) (In solving this problem, the following technique has been employed :



p.21

As 1, x, 3

1 are in a G.S.,

3

13

1

3

1)1(

2

2

x

x

x

(b) As 3

1x , the common ratio is

3

1.

The n th term

1

1

3

1

3

1)1(

n

n

(c) Let a = the first term and r = the common ratio.

2

33

13

33

)13()13(

)13(3

13

3

33

11

31

3

11

11

r

aS

(d) )12(...)5()3()1( nTTTT

p.22

2

)1(

)1(

2

)1(2

)]1(...321[2

)22(...642

2242

1)12(1513

3

1

3

1

3

1

3

1

3

1

3

1...

3

1

3

1

3

1...

3

1

3

11

nn

nn

nn

n

n

n

n

Ex16.4 16. p.19

(a) The area of 111 CBA 1T

4

39

2

3

2

9

60sin)3)(3(2

1

sin))((2

11111111

CBACBBA

(b) As 2A divides 11BA in the ratio 1 : 2,

121 AA and 212 BA

Similary, 121 BB .

In 212 BBA , 60cos)2)(1(2)2()1()( 22222 BA

32

1441

321 BA

As 222111 ~ CBACBA ,

The area of 222 CBA 2T

p.23

4

33

9

3

4

39

3

3

4

392

(c) (i) Let r = the common ratio. (ii) The general term nT

3

14

393

1

4

39

1

2

T

Tr

3

3

21

1

11

34

3

3

1

4

3

3

1

4

3

3

1

4

39

n

n

n

n

nrT

(iii)

3

11

3

1

3

1

4

3

3

11

3

11

3

1...

3

1

3

1

3

1

3

1

4

3

3

1...

3

1

3

1

3

1

3

1

4

3

34

3...

34

3

34

3

34

3

34

3

...

22

31012

31012

31012

321

n

n

n

n

nTTTT

n

n

n

n

n

3

11

8

327

3

2727

8

3

3

127

8

3

23

127

4

3

33

11

33

1

3

1

4

3

3

3

22

(iv) When n tends to infinity, n3

1 tends to zero.

Thus, the sum to infinity of the G.S. 8

327

p.24

Ex16.4 17. p.20

(a) 1007 = 14.2857 9997 = 142.714

157 = 105 1427 = 994

Thus, the smallest and the largest multiples of 7 between 100 and 999 are 105 and 994

respectively.

(b) Consider the sequence 105, 112, 119, … , 994

which can be rewritten as 715, 716, 717, … , 7142

The number of multiples

128

14142

The sum of these multiples

70336

)10048(7

)10510153(7

2

1514

2

1431427

)]14...321()142...17161514...321[(7

)142...171615(7

1427...177167157

994...119112105

(c) The sum of all positive three-digit integers which are NOT divisible by 7

424214

703362

10099

2

1000999

70336)]99...321()999...321[(

70336)999...102101100(

Ex16.4 18. p.21

(a) 121 CC

As 321 CCC is an equilateral triangle, Area of 60sin)1)(1(2

1321 CCC

131 CC and 60213 CCC . 2m

4

3

2

3

2

1

p.25

(b) The area of a smaller equilateral triangle : The area of 321 CCC = 2)3:1(

Thus, the area of a smaller equilateral triangle 9

1

4

3

36

3

The total area of all equilateral triangles 36

34

4

3

2m 36

313

9

41

4

3

4

3

9

4

4

3

(c) Repeat the process in (b) once, the area of the smallest triangle = 4

3

9

12

The total area of all the equilateral triangles 4

3

9

116

9

41

4

32

2

2

2

m 9

4

9

41

4

3

9

4

4

3

9

41

4

3

Thus, after repeating the process many times,

the total area of all the equilateral triangles

...

9

4

9

41

4

32

2m 20

39

49

9

4

3

9

41

1

4

3

Ex16.4 19. p.23

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

p.26

5a

3

58

d

0785073

7850)73(

7850)3310(

3925)]3)(1()5(2[2

3925])1(2[2

3925)(

2

nn

nn

nn

nn

dnan

nS

50n or 3

152n

(rejected)

Thus, 50n .

Ex16.4 20. p.23




(a) As 10, b, 24 is an A.S., bb 2410

17

234

b

b

The sequence is now a, 10, 17, 24.

As a, 10, 17 is an A.S., 101017 a

3a

(b) (i) As the salary tax is paid at the standard rate,

P

P

2.0

%20

(ii)

Net chargeable income ($) Rate

On the first 30 000 3%

On the next 30 000 10%

On the next 30 000 17%

Remainder 24%

p.27

In calculating the least net total income so that the salary tax is charged at the standard rate,

let y = remainder.

Thus, the tax 24.017.0300001.03000003.030000 y

y24.09000

The net total income y 300003000030000172000

y 262000

Tax paid at the standard rate 2.0)262000( y

Put 2.0)262000(24.09000 yy

1085000

90005240004.0

2.05240024.09000

y

y

yy

Thus, the least net total income required = $(262000 + 1085000)

= $1347000

(c) The tax Peter needs to pay

280000$

2.01400000$

The money he saves

3706.28052610025.1

0025.10025.123000

0025.11

0025.10025.123000

)0025.1...0025.10025.10025.1(23000

12

03.0123000...

12

03.0123000

12

03.0123000

12

03.0123000

13

13

1232

1232

Thus, Peter does not have enough money to pay for the tax.

Ex16.4 21. p.25

(a) (i) The taxi fare

)126$(

)241230$(

)12)2(30$(

4.22.0

)2(30$

x

x

x

x

p.28

(ii) When x is not a multiple of 0.2, bxa where both a and b are multiples of 0.2.

The fare for a is a126

and the fare for b is b126

The taxi fare for x will not be x126 . It will be b126 .

(b) 2.31.30.3

The taxi fare

4.44$

)2.3126$(

(c) Starting from the second journey, the sequence of the journeys in km is

3.6, 4.1, 4.6, 5.1, 5.6, 6.1, 6.6, 7.1, … , 51.6, 52.1

which is the combination of two sequences:

Sequence I : 3.6, 4.6, 5.6, … , 51.6

Sequence II : 4.1, 5.1, 6.1, … , … , 52.1

When taxi fares are charged, sequence II becomes sequence III : 4.2, 5.2, 6.2, … , 52.2

The total taxi fare for sequence I

8.16522

)]1)(149()6.3(2[2

4912649

)]6.51(126[...)]6.5(126[)]6.4(126[)]6.3(126[

The total taxi fare for sequence III is

6.16875

)]1)(149()2.4(2[2

4912649

)]2.52(126[...)]2.6(126[)]2.5(126[)]2.4(126[

The total fare

= $( 6.168758.165224.44 )

= $ 8.33442

> $33000

The taxi driver is not right.

Ex16.5 1. p.27

The sum of the multiples of 3 between 1 and 300, including 1 and 300

15150

2

1011003

)100...321(3

300...963

(ANS : C)

p.29

Ex16.5 2. p.27

Let a = the 1st term and r = common ratio.

It is given that 2nd term = 24 and 5th term = 3

24ar … (1) 34 ar … (2)

(2)(1),

2

1

8

1)(

8

124

3

3

1

3

13

3

4

r

r

r

ar

ar

Sub 2

1r into (1), 24

2

1

a

48a

Thus, the first term is 48.

(ANS : C)

Ex16.5 3. p.27

Let a = the 1st term and d = common difference.

The tenth term is three times its second term.

2

126

31896

))6((3)9)6((

)(3)9(

d

d

dd

dd

dada

(ANS : D)

Ex16.5 4. p.28

The sum of the first five terms of an A.S. is 15.

Let the five terms be dy 2 , dy , y , dy , dy 2 .

Thus, 155 y

3y

The fourth term is given to be 7.

The common difference

p.30

4

37

The first term

5

)4(23

2

dy

(ANS : A)

Ex16.5 5. p.28




First, as a, b, c are in G.S., 2bac .

We will make use the fact that 2bac in the following checking.

I. Checking whether 2a , 2b , 2c are in G.S.

22 ca

2)(ac 22 )(b (Using the fact that 2bac )

Thus, 2a , 2b , 2c are in G.S.

II. Checking whether ab , ac , bc are in G.S.

acb

bcab2

acac)( (Using the fact that 2bac )

2)(ac

Thus, ab , ac , bc are in G.S.

III. Checking whether a

1,

b

1,

c

1 are in G.S.

ac

ca1

11

2

1

b (Using the fact that 2bac )

2

1

b

Thus, a

1,

b

1,

c

1 are in G.S.

(ANS : D)

p.31

Ex16.5 6. p.28




First, as a, b, c are in A.S., bca 2 .

We will make use the fact that bca 2 in the following checking.

I. Checking whether 2a , 2b , 2c are in A.S.

1, 2, 3 are in A.S.

1, 4, 9 are not in A.S.

2a , 2b , 2c are not in A.S. generally.

II. Checking whether a6 , b6 , c6 are in A.S.

)(6

66

ca

ca

)2(6 b (Using the fact that bca 2 )

)6(2 b

Thus, a6 , b6 , c6 are in A.S.

III. Checking whether a

1,

b

1,

c

1 are in A.S.

1, 2, 3 are in A.S.

1, 2

1,

3

1 are not in A.S.

a

1,

b

1,

c

1 are not in A.S. generally.

(ANS : B)

Ex16.5 7. p.29

a is the arithmetic mean between )43( ba and )54( ba .

In other words, )43( ba , a , )54( ba are in A.S..

Thus, ababa 2)54()43( (Using 2231 TTTT for an A.S.)

ab

ba

aba

5

5

27

(ANS : D)

Ex16.5 8. p.29

For the G.S. 1, 1.0 , 01.0 , 001.0 , …

p.32

Let a = the 1st term and r = the common ratio.

1a , 1

1.0r

r

aS

1

1.0 )1.0(1

1

1.1

1

11

10

(ANS : D)

Ex16.5 9. p.29

For the G.S. 4, a , b , 25

100

254

ba

(ANS : C)

Ex16.5 10. p.30

Let the angles be y , )20( y , )40( y , )60( y , )80( y , )100( y ,

Sum of the angles 1804

720

70

4206

7203006

720)100()80()60()40()20(

y

y

y

yyyyyy

The largest angle is 170. (ANS : C)

Ex16.5 11. p.30

Referring to Ex16.5 5. p.28, I and II are true.

Check if alog , blog , clog are in A.S.

b

b

ac

ca

log2

)log(

)log(

loglog

2

Thus, III is true.

(ANS : D)

p.33

Ex16.5 12. p.30

Let a = the 1st term and d = the common difference.

Note that 101 TT

92 TT

83 TT

74 TT

65 TT

Thus, 5

120101 TT

6

122

36242

24)4(92

24)9(

a

a

a

a

daa

(ANS : B)

Ex16.5 13. p.31




As a

1,

b

1,

c

1 are in G.S.,

acb

bac

bac

bbca

2

2

2

11

1111

(ANS : A)

Ex16.5 14. p.31

The areas are in the sequence 1, 2

2

1

,

2

4

1

, …

p.34

It is a G.S. with the first term 1a and the common ratio 4

1r .

The sum of the areas ...4

1

2

11

22

3

44

3 1

4

11

11

r

a

(ANS : E)

Ex16.5 15. p.32

Let the common ratio be r.

1

2

T

Tr

a

b

The sum to infinity r

a

1

1

2

ba

a

a

baaa

ba

(ANS : E)

Ex16.5 16. p.32

(You may refer to Ex16.5 12. p.30)

After inserting the four arithmetic means, the sequence becomes

12, 2T , 3T , 4T , 5T , 27

p.35

78

)2712()2712(

)()( 5243

5432

TTTT

TTTT

(ANS : A)

Ex16.5 17. p.32

The quickest way is thinking of a multiple of 8 and it should be as large as possible such that when

2006 minus this number, the result is still positive.

The number is 2000.

Thus, 2006 2000

= 6

When 6 minus 8 again, the sequence has its first negative number which is 2.

(ANS : D)

Ex16.5 18. p.33

(You may refer to Ex16.5 12. p.30 and Ex16.5 16. p.32)

As 1624... 28321 aaaa ,

11614

1624281

aa

Now, as 621 aa ,

21 6 aa

Thus, the common difference d of the arithmetic sequence is 6.

23

462

1161622

116)6(272

116272

116)27(

116

1

1

1

1

1

11

281

a

a

a

a

da

daa

aa

(ANS : C)

p.36

Ex16.5 19. p.33

The sequence 4, 2 , 1, 2

1 ,

4

1,

8

1 , …

has the first term 4 and the common ratio 2

1 .

The sequence 4, 1, 4

1, …

has the first term 4 and the common ratio 2

2

1

.

For the sequence with positive terms only, let a = the first term and r = the common ratio.

3

163

444

3 4 4

11

41

...4

114

r

a

(ANS : C)

Ex16.5 20. p.33

In solving problems of this kind, you are advised to use observation and intuitive understanding

instead of complicated mathematical calculation.

Note that the first row of the “square-like” picture of dots are

3 in the 1st pattern,

4 in the 2nd pattern,

5 in the 3rd pattern

and it is reasonable to assume that the coming ones should be 6, 7, 8, 9.

Therefore the number of dots in the 7th pattern

199

82

(ANS : C)

p.37

Ex16.5 21. p.34

As 6a , a , 5a are in G.S.,

2)5)(6( aaa (Using the fact that 2231 TTTT )

30

030

030

30 22

a

a

a

aaa

The common ratio

6

536

3036

30630

30

(ANS : B)

Ex16.5 22. p.34

First, we calculate the value of 54321 TTTTT .

Then, we calculate the value of 4321 TTTT .

After doing subtraction, we can find the value of 5T .

35

)5(2)5( 2

54321

TTTTT

24

)4(2)4( 2

4321

TTTT

11

24355

T

(ANS : B)

Ex16.5 23. p.34

Let a = the 1st term and r = the common ratio.

As 327 a and 89 a ,

we have 326 ar … (1) and 88 ar … (2)

p.38

(2) (1), 32

86

8

ar

ar

4

12 r

2

1r or

2

1r

Whatever r is, 4

12 r .

Thus, 9a , 7a , 5a , 3a , 1a are all positive. Therefore, I is true.

When 2

1r ,

aa 1 , aa2

12

a

aa

aa

2

12

121

0 (as a is always positive)

When 2

1r ,

aa 1 , aa2

12

a

aa

aa

2

3

2

121

0 (as a is always positive)

Thus, II is true.

1005432 ... aaaaa First a is always positive (see work above). 99432 ... ararararar r1 is always positive either when r is 0.5 or 0.5.

)...( 99432 rrrrra 100rr is positive when r is 0.5

r

rrrrrra

1

)1)(...( 99432

and it is negative when r is 0.5. Thus, III does not

r

rra

1

100

hold.

(ANS : A)

Maths ch16 key

Education

Transcript of Maths ch16 key