3.APGP Series Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 9758 H2 Mathematics 2023 Sequence & Series T1 Solutions (Sequencs and Series) 1(a) First term =1, common difference = d 5 10 20,,S S S form a GP .  10 20 5 10 SS SS        2 10 5 202 9 2 4 . 2 192 2 2 d d d           2 2 9 2 4 . 2 19d d d     25 10 0dd   2d or d = 0 (rejected as AP is increasing)       1 2 2 22 2 2 2 2 1100 2 1 1 100 2 ( 1) 100 0 1 100 2 99 0 1 100 n n S S n n nn n nn n             Since 2 2 99 0nn   as discriminant < 0 and coefficient of n2 is +ve.   2 1 100 0 ( 1 10)( 1 10) 0 ( 11)( 9) 0 9 n nn nn n              least n is 10. (b) nth month Outstanding amt owed at the start of the month (in hundreds) Outstanding amt owed at the end of the nth month (in hundreds) 1 34 34 2 (34)2 (34)2 3 (34)22 34(2)2-70 4   22 34(2) 70 334(2) 70(2) 70 5   32 34(2) 70(2) 70 4234(2) 70(2) 70(2) 70   n 1334(2) 70(2) .......... 70(2) 70nn    Total amount of money owed at the end of nth month

River Valley High School, Mathematics Department 9758 H2 Mathematics 2023 Sequence & Series T2 = 100( 1334(2) 70(2) .......... 70(2) 70nn    ) 2 1 70((2) 1)100 34(2) 21 n n        1100 70 2 n To be free from debt, 170 2 0n 2 140 ln140 7.129ln 2 n n     Least n = 8 Earliest month to be free from debt = Jan 2014 2(i) Length of film in the 1st layer = 42 Length of film in the 2nd layer = (42 + 2x) Length of film in the 3rd layer = (42 + 4x) Length of film in the nth layer = [42 + 2x(n −1)] = 124 2x (n −1) = 82 ( 1) 41xn   (shown) 2(ii) nS = 16766 mm, a = 42 mm, l = 124 mm Using ()2 n nS a l 2 2(16766π) 20242π 124π nSn al   Using (i), x(n – 1) = 41, 41 0.203981x n  x = 0.204 (to 3 s.f.) 2(iii) The sum of the first 10 terms = k (the sum of the first 5 terms) 10 5(1 ) (1 ) 11 a r a r krr   10 51 (1 )r k r   5 5 5(1 )(1 ) (1 )r r k r   

River Valley High School, Mathematics Department 9758 H2 Mathematics 2023 Sequence & Series T3 Since 51, 1 0rr   Therefore, 51kr (shown) 2(iv) Since k =33 and a = 50, 5 5 1 33 32 2 r r r    Suppose the roll of film is cut into n pieces. The sum of the first n terms  total length of the film 50(2 1) 5258021 2 1 1051.6 2 1052.6 ln 2 ln1052.6 10.03974 n n n n n        the largest possible value of n is 10. Hence the largest number of pieces is 10 pieces. 3 22 21 n nU ar    . (i) Given that n n nU U U U U      1 3 2 1 2 2 1 ie n n n na ar ar ar ar ar         2 2 2 2 1 2 2 1    n nar ar rr    2 21 2 1 11    nnrr r r r     2 2 11 1 1 1   nnr r r    2 2 111 nnrr    2 2 12 1 0 [Shown] When n 5 , we have rr   10 92 1 0 So r  1 (NA convergent series) or .0 892 (ii) Now    1 11 n nV ar . Letting the nth term of the new sequence be nT , we have    ln ln lnn n n n n UT U V V    

River Valley High School, Mathematics Department 9758 H2 Mathematics 2023 Sequence & Series T4     ln ln ln ln           111a n r a n r  ln ln  2 2 1a n r    ln ln ln ln      1 2 2 1 2 2 2nnT T a n r a n r ln r 2 , a constant Hence the new sequence is an arithmetic progression Alternative:               ln ln ln ln ln ln ln ln ln ln constant                      n n n n n n n n n n n n n n UT U V U U UV arT T U U r ar 1 11 2 2 2 2 2 2 4 (i) Total volume: 850 m3 Volumes (in m3) filled: 4, 4.5, 5, 5.5, 6, … … Assume that the container is completely filled at the nth pumping action. For the AP above, ----- (*) Considering we have , i.e., (or use G.C.) Hence from (*), at the 52th pumping action, the container would be completely filled, with some liquid overflowing. volume of liquid that overflows m3 Alternative Using G.C.,  2(4) ( 1)(0.5) 8502 n nSn    2 15 3400 0nn   2 15 3400 0,nn   215 15 4(1)( 3400) 2(1)n     51.2899 or 66.2899nn  ( 51.2899)( 66.2899) 0nn   66.2899 (NA, since 0) or 51.2899n n n      52 52 2(4) 51(0.5)2S  871    3871 850 m 21  2(4) ( 1)(0.5) 8502 n nSn   

River Valley High School, Mathematics Department 9758 H2 Mathematics 2023 Sequence & Series T5 n 51 52 53 841.5 871 901 volume of liquid that overflows m3 (ii) For machine A, Volumes (in m3) filled by B: For the GP above, total volume filled after 10th pumping action by B (b) (For the GP above, since sum to infinity exists.) Sum to infinity Volume of container to be filled by machine B Since the theoretical maximum volume that B alone can fill is only 30 m3, which is less than 37.5 m3, the container would never be completely filled.  2(4) ( 1)(0.5)2 n n    3871 850 m 21  50 50 2(4) 49(0.5)2S  812.5 2 555, 5 , 5 , ...66            10 10 551 6 51 6 S      3156941 25.1548 m6239  3(812.5 25.1548) m 3837.6548 m 3838 m 5| | 1, 6r  5 51 6   30   33850 812.5 m 37.5 m  

River Valley High School, Mathematics Department 9758 H2 Mathematics 2023 Sequence & Series T6 Alternative: Since the container can only be filled to a theoretical maximum total volume of 812.5 + 30 = 842.5 m3 , which is less than 850 m3, the container would never be completely filled. 5(a) (i) Note that the progression of A is 1,3,5,7,9,11, … (consists of all positive odd numbers) nth term of G, nT  132 na   132 1na =  1332  nna =  13)3(2 1 na = )3( 1na 13 n is odd for all 1n . Hence, for )3( 1na to be a term in A, then a = 12 m ,  Zm . That is, a may be any positive odd integer. (ii) Let the new progression be H : 12t , 22t , 32t , 42t , … Let d be the common difference of A. Consider, H H of th term)( of th term)1( n n = ntnt nt nt n n H H    1 11 2 2 2 = 422 2 d , which is a constant. Hence H is a geometric progression. (b)(i) Value of the machine after 5 years = 5)9.0(70000 = $41334.30. (ii) Maximum revenue that could possibly be generated by the machine = 200000$)93.0(1 14000  . Consider at the nth year of operation, Total revenue + prevailing value − cost of new machine  40000 4000070000)9.0(70000)93.0(1 ))93.0(1(14000   n n 47)9.0(7))93.0(1(20  nn 9)9.0(7)93.0(20  nn From G.C., n na … … 5 9.7803 6 9.2197 7 8.6859  7n Hence, the number of years the machine is in operation is 7.

River Valley High School, Mathematics Department 9758 H2 Mathematics 2023 Sequence & Series T7 6 7(i)(a) Total amount paid = 600 + 650 + 700 + ….       2 600, 50 2 600 1 502 600 25 25 25 575 (Shown) n ad nSn nn nn         Total paid after nth payment   2$ 25 575nn (i)(b) To complete fully repay her study loan, 2 Amount paid 30000 2500 25 575 32500 0 From GC, 49.345 or 26.345 nn nn       It will take Emma 27 payments to fully repay her study loan. (ii)(a) Mth Amount owed at start of the month Amount owed at end of the month Jan (n=1) 30000 x 30000 x

River Valley High School, Mathematics Department 9758 H2 Mathematics 2023 Sequence & Ser