Chapter 7: Arithmetic and Geometric Series (Answer Sheet)
Q1.
“Sum of al the terms in the geometric progression = sum of all the terms in arithmetic progression”
à First term of geometric progression is 192 thus a = 192
à Common ration is 1.5 so r = 1.5
à 6 term so n = 6
( ) à Sum of all terms in geometric progression = � =
( . ) à � = .
à 3990
à Common difference is 1.5 so d = 1.5, 21 terms thus n = 21 however we do not know the first term of the arithmetic progression.
à Sum of all terms in arithmetic progression = � = [2� + � − 1 �]
à � = [2� + 21 − 1 × 1.5]
à 21a + 315
à 21a + 315 = 3990
à a = 175, 175 is the first term of the arithmetic progression
à The last term of the arithmetic progression
à 175 + 21 − 1 × 1.5 = 205
à 175, 205
Q2. i)
Second term of geometric progression is 3 thus �� = 3
à �� = 3
à Sum to infinity is 12
à � = = 12
à � = 12(1 − �)
à � = 12 − 12�
à Substitute a in �� = 3 with � = 12 − 12�
à �� = 3
à (12 − 12�)� = 3
à 12� − 12� − 3 = 0
à −4� + 4� − 1 = 0
à −2� + 1 2� − 1 = 0
à � =
à Input � = into �� = 3
à � = 3
à � = 6
à 6 ii)
First term of the geometric progression is 6 and the second term is 3, so � of the arithmetic progression is 6 and � is 3.
à � = � + 2 − 1 �
à 3 = 6 + �
à � = −3
à Find the sum of the first 20 terms with n = 20
à � = [2(6) + 20 − 1 × (−3)]
à −450
Q3. i)
First term of geometric progression is 81, so a = 81
à Fourth term is 24, thus �� = �� = 24
à Input a = 81 into �� = 24 to find the value of common ratio r
à 81� = 24
à � =
à � = = ii)
a = 81 , � =
à Sum to infinity = � =
à = 243 iii)
Second term of geometric progression �� = �� = 81 × = 54
à Third term = �� = �� = 81 × = 36
à First term of arithmetic progression becomes 54 and the fourth term becomes 36
à Find the common difference
à � = 54
à � = � + 4 − 1 �
à 36 = 54 + 3�
à −18 = 3�
à � = −6 à Find the sum of the first ten terms
à � = 2 54 + 10 − 1 × −6 = 270
Q4. a)
� = 0.5, �� = �� = 0.5 , �� = �� = 0.5
à Common ratio can be found by either inputting a = 0.5 into �� = 0.5 or simply by dividing the second term by the first term
à In the geometric progression, the second term is divided by the first term (� ÷ � ) always gives the common ratio
à �� = 0.5 , � = 0.5
à 0.5� = 0.5
à � = 0.5 = 0.25
à Sum to infinity
. à � = = = . b)
� = 5, � = 9
à Get the common difference
à � = � + 2 − 1 �
à 9 = 5 + 2 − 1 �
à 9 = 5 + �
à � = 4
à Now find the number of terms
à The last term has to be the only term that is greater than 200
à 5 + � − 1 4 > 200
à � − 1 4 = 195
à � = 49.75
à Rounding up 49.75 to the nearest integer becomes 50
à Test with n = 50 and n = 49
à 5 + 49 − 1 × 4 = 197 while 5 + 50 − 1 × 4 = 201
à 201 is bigger than 200 thus the 50th term is the last term in this progression
à Find the sum of 50 terms
à � = 2 5 + 50 − 1 × 4 = 5150
Q5. a) Multiples of 5 such as 5, 10, 15, 20… have a common difference of 5, thus d = 5
à The question is asking for the sum between 100 and 300 inclusive, thus the first term is 100
à a = 100
à Find the number of terms from the fact that 300 is the last term
à 300 = 100 � − 1 5
à 300 = � − 1 5
à 40 = � − 1
à � = 41
à Sum of 41 terms
à � = 2 100 + 41 − 1 5
à 8200 b) i)
Common ratio � = −
à Sum of the first 3 terms is 35
( ) ( ) à � = = = ( )
à = 35 à � = 35 ×
à � =
à � = 45
à The first term of the progression is 45 ii)
Sum to infinity
à � = = = 27 ( )
Q6. i)
Be aware that the question is asking for the amount of donation made not the amount of prizes are 40 days.
à Model 1 uses the arithmetic sequence and has a common difference of $1000 starting from the first term of $1000 on day 1.
à The donation is 5% of the value of the prize on the day, thus for prizes of $1000, $2000, $3000… the donation will be 1000 × 0.05 = $50 , 2000 × 0.05 = $100 , 3000 × 0.05 = $150 , 4000 × 0.05 = $200
à The donation made has an arithmetic sequence with the first term $50 and a common difference of $50.
à Fid the total amount donated to charity for 40 days
à � = 2 50 + 40 − 1 × 50 = $41000 ii)
Part (i) was solved by finding the sequence of the donations made each day, then getting the sum after 40 days, however you can also find the sum of prizes over 40 days then get 5% of the sum
à Model 2 uses a geometric sequence where the prize increases by 10% each day so the common ratio will be 100%+10%
à +
à 1 + 0.1
à � = 1.1
à First term is $1000
à Find the sum of prizes over 40 days
( . ) à � = = 442592.5557 .
à 5% of the sum of prizes was made for charity donation
à 442592.5557 × 0.05
à 22129.62778
à $22100
Q7. a)
The angles of the sectors are in arithmetic progression and the sum of all angles of all 6 sectors is 360° as they form a circle
à The angle of the largest sector is 4 times the angle of the smallest sector
à Smallest sector’s angle = a , angle of the largest sector = 4a
à Since there are 6 sectors, the angle of the largest sector will be the 6th term
à � = � + 6 − 1 �
à 4� = � + 5�
à 3� = 5�
à � = �
à Sum of all angles = 360°
à � = 2 � + 6 − 1 �
à 360° = 3 2� + 5�
à 360° = 6� + 15�
à Input � = � into 360° = 6� + 15�
à 360° = 6� + 15 �
à 360° = 6� + 9�
à 360° = 15�
à � = 24°
à The angle of the smallest sector = 24°
à Convert to radians
à24 × = � °
à Perimeter of sector = radius + radius + arc
à Radius = 5cm
à Arc = 5 × � = �
à Perimeter = 5 + 5 + �
à 12.1 b) i)
Second term÷ first term = common ratio and third term ÷ second term = common ratio (� ÷ � = �)
à �� ÷ � = � , �� ÷ �� = �
à � + 6 ÷ 2� + 3 = � , � ÷ � + 6 = �
à =
à � + 6 � + 6 = �(2� + 3)
à � + 12� + 36 = 2� + 3�
à � − 9� − 36 = 0
à � − 12 � + 3 = 0
à � = 12, −3
à All the terms of the geometric progression are positive thus k has to be positive
à k = 12 ii)
� = 2� + 3 = 24 + 3 = 27
à � = = = = ×
à � = = = 81
Q8. a) a = 1 and the common difference is “second term – first term � − � ”, thus cos � − 1
à Find the sum of the first ten terms
à � = 2 1 + 10 − 1 ×(cos � − 1) = 5 2 + 9 cos � − 9
à 5 cos � − 7 = 45 cos � − 35
à Convert cos � to sin � as 1 − sin � = cos �
à 45(1 − sin �) − 35
à 45 − 45 sin � − 35
à 10 − 45 sin � b) i)
For geometric progression to be convergent, the progression has to have a sum to infinity, thus the condition “ � < 1” has to be met.
à The common ratio is the second term ÷ first term
à � = tan � ÷ 1 = tan �
à Now meet the condition
à tan � is always a positive number thus 0 < tan � < 1
à 0 < tan � < 3
à 0 < tan � < 3
à tan (0) < � < tan ( 3)
à 0 < � < ii)
Find the sum to infinity with � = 1 , � = tan � , � = �
à � = =
à
Q9. a)
� = � + 8�
à When n = 1 then the value � will just be the value of the first term as � is just the sum of the first term.
à � = 1 + 8 × 1 = 9
à The value of the first term is 9
à a = 9
à When n = 2 then the value of � will be the sum of the first term and the second term
à � = � + �
à � = 2 + 8 × 2 = 20
à 20 = � + �
à 20 = 9 + �
à � = 11
à The common difference will be the second term – the first term, � − �
à 11 − 9 = 2
à a = 9 , d = 2 b)
“second term is 9 less than the first term”
à a – 9 = ar
à a – ar = 9
à a(1 –r ) = 9
à � =
à “sum of the second and third term is 30”
à �� + �� = 30
à Input � = into �� + �� = 30
à � + � = 30
à 9� + 9� = 30(1 − �)
à 9� + 9� = 30 − 30�
à 9� + 39� − 30 = 0
à 3� + 13� − 10 = 0
à 3� − 2 � + 5 = 0
à � = , −5
à All the terms are positive thus � =
à Input � = into a – ar = 9
à � − � = 9
à � = 9
à � = 27
Q10. i)
Sum of the first 200 terms = 2� + 200 − 1 �
à 100 2� + 199�
à 200� + 19900�
à Sum of the first 100 terms = 2� + 100 − 1 �
à 50 2� + 99�
à 100� + 4950�
à Sum of the first 200 terms = 4 × the sum of the first 100 terms
à 200� + 19900� = 4 × (100� + 4950�)
à 200� + 19900� = 400� + 19800�
à 100� = 200�
à � = 2� ii)
First term = a , common difference = 2a
à 100th term
à � = � + 100 − 1 × 2� = � + 198�
à 199a
Q11. a) a = −2222 , d = 17
à Find the value of n that will give the first positive term
à −2222 + � − 1 17 > 0
à −2222 + 17� − 17 > 0
à 17� − 2239 > 0
à 17� > 2239
à � > 131.706
à Rounding up the n to the nearest integer gives n = 132
à Test with n = 131 and n = 132
à For n = 131, −2222 + 131 − 1 17 = −12 thus negative number
à For n = 132, −2222 + 132 − 1 17 = 5
à 132nd term gives the first positive term
à 5 b)
� = 3 , �� = 2 cos �
à Find the value of r
à �� ÷ � = �
à 2 cos � ÷ 3
à � =
à For the progression to be convergent, the progression should meet the condition that � < 1 which is −1 < � < 1
à −1 < < 1
à − 3 < 2 cos � < 3
à < cos � <
à The value of � lies in 0 < � < �
à cos = � , then for simply find the angle in the 2nd quadrant so that � − � = �
à � < � < �
à cos = �
Q12. i)
Plan A follows a geometric
à Profit increases each year by 5%, thus the common ratio is 100% + 5% = 105%
à r = 105% = = 1.05
à The first term a = $250000
à Find the profit of year 2008
à Since year 2000 is the first term, then year 2008 will be the 9th term.
à ��
à 250000 × 1.05 = 369363.8609
à $369000 ii)
Year 2009 will be the 10th term
. à .
à 3144473.134
à $3140000 iii)
Plan B follows an arithmetic progression with a common difference of $D
à Find the sum for 10 years
à 2 250000 + 10 − 1 × �
à 5(500000 + 9�)
à 2500000 + 45�
à Has to be equal to the answer in (ii)
à 2500000 + 45� = 3144473.134
à 45� = 644473.1339
à � = 14321.6252
à $14300
Q13. i)
� = � + 5 − 1 �
à � = � + 4�
à � = � (15 − 1)�
à � = � + 14�
à a + 4d , a + 14d ii)
First term of geometric progression becomes a, second term is a + 4d and the third term is a + 14d
à In a geometric progression, � ÷ � = � , thus second term ÷ first term and third term ÷ second term will both give the common ratio
à =
à � + 4� � + 4� = �(� + 14�)
à � + 8�� + 16� = � + 14��
à 8�� + 16� = 14��
à 16� = 6��
à 16� = 6�
à 8� = 3� iii)
3� = 8�
à = �
à Input � = to
à = =
à 2.5
Q14. a)
� = 161 , � = 154
à � = � + �
à 154 = 161 + �
à � = −7
à Now that we know the first term and the common difference, we can find the sum of the first m terms
à � = 2 161 + � − 1 × (−7) = 322 − 7� + 7 = 329 − 7�
à This sum has to equal zero
à 329 − 7� = 0
à � 329 − 7� = 0
à � = 0, 47
à � = 0 doesn’t make sense as the progression cannot have zero number of terms
à � = 47 b) a , r , n
( ) à Sum of the first n terms = � =
à Sum to infinity = � =
à The sum of the first n terms is less than 90% of the sum to infinity
( ) à < ×
( ) . à <
à � 1 − � < 0.9�
à 1 − � < 0.9
à 0.1 − � < 0
à 0.1 < �
Q15. a) a = 100, � = 2000
à � = =
à = 2000
à = 20
à 1 = 20 − 20�
à −19 = 20�
à � =
à Second term is �� = ��
à 100 × = 95 b) i)
� = 90 , � = 80
à Find the value of a and d
à � = � + 3 − 1 � = � + 2�
à � = � + 5 − 1 � = � + 4�
à 90 = � + 2� , 80 = � + 4�
à Simultaneous equation
à −10 = 2�
à � = −5
à Input � = −5 to either equation to get a
à 80 = � + 4 −5 = � − 20
à � = 100
à � = 100 , � = −5 ii)
Sum of the first m terms
à � = 2 100 + � − 1 (−5) = 200 − 5� + 5 = (205 − 5�)
à Sum of the first m + 1 terms
à � = 2 100 + � + 1 − 1 (−5) = (200 − 5�)
à These two sums have to be equal
à 200 − 5� = (200 − 5�)
à � 200 − 5� = (� + 1)(200 − 5�)
à 205� − 5� = 200� − 5� + 200 − 5�
à 205� = 200� + 200 − 5�
à 10� = 200
à � = 20
à This question can easily be solved another way
à Since the sum of the first m terms and (� + 1) are equal, that means the � + 1 term does not change the value of the sum
à � + 1 term is thus zero
à � = 0
à 0 = � + � + 1 − 1 � = 100 + �(−5)
à 5� = 100
à � = 20
à � = 20 iii)
Sum of the first n terms is zero
à � = 2 100 + � − 1 (−5)
à (200 − 5� + 5)
à (200 − 5�)
à 200 − 5� = 0
à � 200 − 5� = 0
à � = 0, 41
à � = 0 doesn’t make sense as the progression cannot have zero terms
à � = 41
Q16. a)i)
First term is −15
à Find the sum of 25 terms to get the common difference
à � = 2 −15 + 25 − 1 �
à 2 −15 + 25 − 1 � = 525
à −30 + 24� = 42
à 24� = 72
à � = 3
à � = 3 ii)
The last term is the 25th term
à � = −15 + 25 − 1 3 = −15 + 24 × 3 = 57 iii)
Positive terms start from 0 as 0 , 3 , 6 , 9… until 57
à There are 20 terms from 0 to 57
à � = 2 0 + 20 − 1 (3) = 10 0 + 19 × 3 = 570 (there are other methods of solving this too) b) i) a = 4000 and the common ratio is 100% + 5% thus 1.05
à Starting from 2012 then 2022 will be the 11th term.
à �� = �� = 4000 × 1.05 = 6515.578507
à $6520 ii) n = 11 as 2012 ~ 2022
( . ) à � = = 56827.14865 .
à $56800
Q17. a)
Sum of the first 10 terms
à � = 2� + 10 − 1 � = 5 2� + 9� = 10� + 45�
à 10� + 45� = 400
à 2� + 9� = 80
à “sum of the first 20 terms – sum of the first 10 terms = sum of the next 10 terms”
à � − 400 = 1000
à 2� + 20 − 1 � − 400 = 1000
à 10(2� + 19�) − 400 = 1000
à 2� + 19� − 40 = 100
à 2� + 19� = 140
à Carry out simultaneous equations with 2� + 9� = 80 and 2� + 19� = 140
à 10� = 60
à � = 6
à Input d = 6 into either equation to get a
à 2� + 9 6 = 80
à 2� + 54 = 80
à 2� = 26
à � = 13
à d = 6 , a = 13 b)
For the first geometric progression � = � , � = � , � = 6 thus = 6
à � = 6(1 − �)
à � = 6 − 6�
à For the second geometric progression � = 2� , � = � , � = 7 thus = 7
à 2� = 7(1 − � )
à 2� = 7 − 7�
à Input � = 6 − 6� into 2� = 7 − 7�
à 2 6 − 6� = 7 − 7�
à 12 − 12� = 7 − 7�
à 7� − 12� + 5 = 0
à 7� − 5 � − 1 = 0
à � = , 1
à r = 1 doesn’t make sense since the geometric progression will all have the same terms and have no sum to infinity is the common ratio is 1
à Input � = to any equation to get the value of a
à � = 6 − 6�
à � = 6 − 6 =
à � = , � =
Q18. a)i)
For the first mile, athletes takes 5 minutes, thus 300 seconds
à Following miles take 12 seconds longer than the preceding mile, thus the common difference is 12 seconds
à a = 300 , d = 12
à nth mile takes 9 minutes thus 540 seconds
à �� = � + (� − 1)�
à 540 = 300 + (� − 1)(12)
à 240 = (� − 1)(12)
à 20 = � − 1
à n = 21 ii)
26 miles, thus n = 26
à Find the sum of time taken to run 26 miles
à � = 2(300) + 26 − 1 (12)
à 13 600 + 25 (12)
à 11700 seconds
à 11700 ÷ 60 = 195 minutes
à 195 ÷ 60 = 3
à 3 hours 15 minutes
à 195 ÷ 60 =
à = 3
à 3 = 3 hours
à = thus 15 minutes b)
�� = 48 , �� = 32
à Get the value of r
à = �
à = = �
à Find a
à �� = 48
à � = 48
à a = 72
à Find the sum to infinity
à � = = = 216
Q19. a)
When n = 1, then � will be just the sum of the first term, thus the value of � will be just the value of the first term
à � = 32 × 1 − 1 = 31
à � = 31
à When n = 2, the value of � will be the sum of the first term and the second term � = 32 × 2 − 2 = 60
à 60 = � + �
à 60 = 31 + �
à � = 29
à � = � + �
à 29 = 31 + �
à � = −2
à � = 31 , � = −2 b)
Sum to infinity is 20
à � = = 20
à � = 20(1 − �)
à � = 20 − 20�
à 20 = 20 − �
à � =
à The sum of the first two terms is 12.8
à � + �� = 12.8
à Input � = into � + �� = 12.8
à � + = 12.8
à20� + 20� − � = 256
à −� + 40� − 256 = 0
à (−� + 8)(� − 32) = 0
à � = 8, 32
à If a = 32 then the geometric progression is invalid since for � = , when a = 32 then r = − , however the question indicated that all the terms are positive, thus the common ration cannot be a negative number
à � = 8
Q20. i)
The question says the sum to infinity of P is the first term, sum to infinity of Q is the second term and sum to infinity of R is the third term of the arithmetic progression
à Sum to infinity of P with a = 2 and r = 1÷ 2 =
à � = = = 4
à Sum to infinity of Q with a = 3 and r = 1÷ 3 =
à � = = = 4.5
à 4 is the first term and 4.5 is the second term of the arithmetic equation
à � = 4 , � = 4.5
à � = � + �
à 4.5 = 4 + �
à � = 0.5
à Find the third term
à � = � + 2� = 4 + 2 × 0.5 = 5
à Sum to infinity of R is the third term thus the sum to infinity of R is 5 ii)
First term of R is 4 and the sum to infinity is 5
à Find the common ratio
à � = = = 5
à 4 = 5(1 − �)
à 4 = 5 − 5�
à 5� = 1
à � =
à Find the sum of the first three terms of R
( ) ( ) à � = = = 4.96