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 = � = # &'( &,-(&'&./0) 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 A &'( 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 -D à � = [2(6) + 20 − 1 × (−3)] -D - à −450 Q3. i) First term of geometric progression is 81, so a = 81 à Fourth term is 24, thus ��E'& = ��F = 24 à Input a = 81 into ��F = 24 to find the value of common ratio r à 81�F = 24 -E à �F = H& I -E - à � = = H& F ii) - a = 81 , � = F $ à Sum to infinity = � = A &'( H& à J = 243 &' I iii) - Second term of geometric progression ��-'& = �� = 81 × = 54 F - - à Third term = ��F'& = ��- = 81 × = 36 F à First term of arithmetic progression becomes 54 and the fourth term becomes 36 à Find the common difference à � = 54 à �E = �& + 4 − 1 � à 36 = 54 + 3� à −18 = 3� à � = −6 à Find the sum of the first ten terms &D à � = 2 54 + 10 − 1 × −6 = 270 &D - Q4. a) � = 0.5, ��-'& = �� = 0.5F, ��F'& = ��- = 0.5F à Common ratio can be found by either inputting a = 0.5 into �� = 0.5F 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 �� = 0.5F , � = 0.5 à 0.5� = 0.5F à � = 0.5- = 0.25 à Sum to infinity $ D./ - à � = = = A &'( &'D.-/ F 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 /D à � = 2 5 + 50 − 1 × 4 = 5150 /D - 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 E& à � = 2 100 + 41 − 1 5 E& - à 8200 b) i) - Common ratio � = − F à Sum of the first 3 terms is 35 NJ I P ) $(&' ) $O $ $(&'( ) I JQ à �F = = NJ = R &'( &'( ) I I IR $ JQ à R = 35 I F/ / à � = 35 × -S F F/ &S/ à � = -S F à � = 45 à The first term of the progression is 45 ii) Sum to infinity $ E/ à �A = = NJ = 27 &'( &'( ) I 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 ED à � = 2 50 + 40 − 1 × 50 = $41000 ED - 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% &DD &D à + &DD &DD à 1 + 0.1 à � = 1.1 à First term is $1000 à Find the sum of prizes over 40 days &DDD(&.&VW'&) à � = = 442592.5557 ED &.&'& 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� F à � = � / à Sum of all angles = 360° + à � = 2 � + 6 − 1 � + - à 360° = 3 2� + 5� à 360° = 6� + 15� F à Input � = � into 360° = 6� + 15� / F 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 Y - à24 × = � ° &/ à Perimeter of sector = radius + radius + arc à Radius = 5cm - - à Arc = 5 × � = � &/ F - à Perimeter = 5 + 5 + � &/ à 12.1 b) i) Second term÷ first term = common ratio and third term ÷ second term = common ratio (�#O& ÷ �# = �) à �� ÷ � = � , ��- ÷ �� = � à � + 6 ÷ 2� + 3 = � , � ÷ � + 6 = � \O+ \ à = -\OF \O+ 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 \O+ &-O+ &H - à � = = = = -\OF -×&-OF -S F $ -S à �A = = J = 81 &'( &' I 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 &D à � = 2 1 + 10 − 1 ×(cos- � − 1) = 5 2 + 9 cos- � − 9 &D - 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- � F F à Now meet the condition & à tan- � is always a positive number thus 0 < tan- � < 1 F à 0 < tan- � < 3 à 0 < tan � < 3 à tan'&(0) < � < tan'&( 3) Y à 0 < � < F ii) & & Find the sum to infinity with � = 1 , � = tan- � , � = � F + & & à �A = g = g g &' hijJ k &' hijJ Y I I 0 , à H 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 F - à All the terms are positive thus � = F - à Input � = into a – ar = 9 F - à � − � = 9 F & à � = 9 F à � = 27 Q10. i) -DD Sum of the first 200 terms = 2� + 200 − 1 � - à 100 2� + 199� à 200� + 19900� &DD 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 �&DD = � + 100 − 1 × 2� = � + 198� à 199a Q11.