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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 .
Let a = first term and d = common difference.
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 .
Let a = first term and d = common difference.
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
The difference between 20 and 40
= 40 20
= 20
Let d = the common difference between two terms.
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
The difference between 27 and 5
= 27 5
= 22
Let d = the common difference between two terms.
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
For the G.S., let a = the first term and r = the common ratio.
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
For the G.S., let a = the first term and r = the common ratio.
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
For the G.S., let a = the first term and r = the common ratio.
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
For the G.S., let a = the first term and r = the common ratio.
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
Let a = first term and d = common difference.
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)
can be rewritten as (999+3), (999+6), (999+9), … , (999+999)
can be rewritten as (999+31), (999+32), (999+33), … , (999+3333)
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)
Therefore, there are 251 terms in the A.S. 1000, 1004, 1008, … , 2000 which are divisible by 4.
(c) For the A.S., 1008, 1020, 1032, … , 1992
can be rewritten as (996+121), (996+122), (996+123), … , (996+1283)
Therefore, there are 83 terms in the A.S. 1008, 1020, 1032, … , 1992 which are divisible by 12.
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
For the G.S., let a = the first term and r = the common ratio.
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 :
2231 TTTT for an A.S.
2231 TTTT for a G.S.)
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
(In solving this problem, the following technique has been employed :
2231 TTTT for an A.S.
2231 TTTT for a G.S.)
(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
(In solving this problem, the following technique has been employed :
2231 TTTT for an A.S.
2231 TTTT for a G.S.)
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
(In solving this problem, the following technique has been employed :
2231 TTTT for an A.S.
2231 TTTT for a G.S.)
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
(In solving this problem, the following technique has been employed :
2231 TTTT for an A.S.
2231 TTTT for a G.S.)
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)