DHS Sequences & Series (9758) (Revision Solutions)

Y5 Topical Revision Package 1 5. Sequences and Series (solutions) 1 5. Sequences and Series (solutions) 1 MJC PROMO 2008/QN6 (a) 2 2 9 , 1< <1 (1)21 2 (2) (2) 4 4: (1 ) 0(1) 9 9 41(n.a as 1) or 33 aSr r T ar r r r r r r r  = = − −−−−−−− = =− −−−−−−−−− − =−  − − = − =  = (b) Method 1 1Sum of first terms, [2 ( 1) ]. ( )2 The last terms are ( ) , ( 1) ,..., ( 1) . I.e. it forms another AP with first term ( ) , last term ( 1) number of terms k n k kk S a k d T a n k d k a n k d a n k d a n d a n k d a n d k −+= + − = + − + − + − + + − = + − = + − = Let ' be the sum of the last terms. ' [ ( ) ( 1) ]2 [2 ] [2 2 ]22 k k Sk kS a n k d a n d kk a nd kd nd d a nd kd d = + − + + − = + − + − = + − − ' [2 2 ] [2 ( 1) ]22 [2 2 2 ]2 [2 2 ] ( )2 kk kkS S a nd kd d a k d k a nd kd d a kd d k nd kd n k kd − = + − − − + − = + − − − − + = − = − Method 2 2 Sum of first terms, [2 ( 1) ].2 Let ' be the sum of the last terms ' [2 ( 1) ] [2 ( 1) ]22 1 [2 2 ]2 k k k n n k kk S a k d Sk S S S n n ka n d a n k d ak knd k d kd − = + − =− −= + − − + − − = + − −

Y5 Topical Revision Package 1 5. Sequences and Series (solutions) 2 2 22 2 1' [2 2 ] [2 ( 1) ]22 1 [2 2 2 ]2 1 [2 2 ] ( )2 kk kS S ak knd k d kd a k d ak knd k d kd ak k d kd knd k d n k kd − = + − − − + − = + − − − − + = − = − Method 3 Sum of first terms, [2 ( 1) ].2 (Think of the last terms as an AP going backwards from the last term.) ( 1) , ( 2) , ( 3) ..., ( ) . Therefore, first term = ( 1) common difference = k kk S a k d k a n d a n d a n d a n k d a n d = + − + − + − + − + − +− number of terms = Let ' be the sum of the last k terms. ' {2[ ( 1) ] ( 1)( )}2 ' {2[ ( 1) ] ( 1)( )} [2 ( 1) ]22 [2 2 ]2 () k k kk d k S kS a n d k d kkS S a n d k d a k d k nd d kd d kd d n k kd − = + − + − − − = + − + − − − + − = − − + − + =− Method 4 Difference between the sum of last k terms and the sum of 1st k terms           ( ) ( ) ( ) 11 1 1 1 ( 1) ( 1) ( 1) ( 1) ( 1) 1 ( ) 1 12 2 2 nk r n k r n n k k r r r a r d a r d a r d a r d a r d n n k kan d n n a n k d n k n k ak d k k = − + = − = = = = + − − + − = + − − + − − + −  −        = + + − − − + + − − + − + + −                     

Y5 Topical Revision Package 1 5. Sequences and Series (solutions) 3 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 2 1 1 12 2 2 1 1 12 2 2 112 22 222 (shown) n n k kd n n n k n k k k n n k kd n n k k d n n n n kn k n k k d kn k n k d kn k dk n k −= + − − + − + − − + + −= + − + − − + = + − − + + + − − − = + − =−  =− 2 HCI PROMO 2010/QN6 (a) Sum of first 3 terms = Sum of the next 6 terms 3 3 6 33 10 aa d a d d+ = +  =− (b) (i) Given ( )21 n nSa − = − − , ( ) ( ) ( ) ( ) ( ) 1 1 121 1 3 2 2 2 2 n n n n n n n a aT S S a a a a − − −− −= − = − = = − − − − ( ) ( ) 1 3 2 3 1 2 1 ,2 n n a an a n a T Ta − − − − − − == − a constant. Thus the sequence is a GP, with common ratio 1 2a− . (ii) For S to exist, 1 1 2 1 3 or 12 a a aa   −    −  : 3 or 1 a a a   3 SAJC PROMO 2010/QN5a (i) ( ) ( ) 1 1 11 22 11! 1 ! 2 2 2 2 2 (2 ) ! 1 ! ! ! nn n n n n n n n n T S S nn nn n n n n + − ++ = − = − − + − −−= − = = − 4 5 8 8 3 94 ... 22 118! 3! 4 8 836 315 3 315 T T T S S+ + = − = − − + = − =−

Y5 Topical Revision Package 1 5. Sequences and Series (solutions) 4 4 JJC PROMO 2009/QN3 (a) (i) 2009a= , 5 7r=− 1 2009 nU  1 512009 7 2009 n− − 1 512009 7 2009 n−   ( ) 2 511 ln ln7 2009n    −        ( ) ( )2ln 20091 5ln 7 n −−  46.2n Least n = 47 (ii) The negative terms of the series is U2, U4, U6, … 35 5 5 52009 , 2009 , 2009 ,...7 7 7      − − −           New GP with first term 52009 14357 − =− , Common ratio 2 5 25 7 49 −= So sum to infinity exists and 1435 2929.79251 1 49 aS r  −= = =−− − (b) 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5………., k…… 12 3 4 5 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... .... kT kTk=  kT is an AP with 1ad== and kS gives the position of the last term the number k appears in the given sequence. Consider 1000kS = ( )1 10002 k k + = 2 2000 0kk + − = 44.2k=  The 1000th term is 45

Y5 Topical Revision Package 1 5. Sequences and Series (solutions) 5 5 HCI PROMO 2009/QN8 (i) 1 1 2 1 14 (a constant)1 4 4 It's a GP. nn n n u u − − − ==  (ii) 111 4 11 4 n nS −= − 4113 4n =−  (iii) 14 1 31 4 S == − or 4 1 4133 4lim nn S → = − =  (iv) , 0.01 Since 4 4 1 41 0.013 3 4 3 1 0.014 n n n n S S S SS   −      − −            ln 0.01 3.32ln 0.25n= Alternatively, Use GC table n = 3, 1 0.01563 0.014 n  = n = 4, 1 0.00391 0.014 n  = least 4n= 6 RVHS PROMO 2009/7 (i) In 2001, 9 (1200) 100 118010 += In 2002, 9 (1180) 100 116210 += (ii) In 2002, 299 (1200) 100 10010 10     ++       In 2003, 329 9 9(1200) (100) 100 10010 10 10      + + +           nth year,

Y5 Topical Revision Package 1 5. Sequences and Series (solutions) 6 12 12 9 9 9(1200) (100) 100 ... 10010 10 10 9 9 9 9(1200) 100 ... 110 10 10 10 n n n n n n −− −−       + + + +                  = + + + + +                 919 10(1200) 100 910 1 10 9 (1200) 1000(1 0.9 ) (Shown)10 n n n n  −  =+  −  = + − (iii) 9As , 0, 10 n n → → Hence population 1000.→ 7 MJC PROMO 2015/QN11 (a) ( ) ( ) ( ) 1 4 1 1 4 1 1 85 n n nT S S n n n n n −=− = − − − − −  =− ( ) ( ) 1 8 5 8 1 5 8 constant nnT T n n−− = − − − − = The series is arithmetic. (b) (i) 2 5 7,,T T T of an AP forms 3 consecutive terms of a GP 2 2 2 2 2 46 4 8 16 7 6 10 0 since 0, 10 10 4 2common ratio, 10 3 2Since 1, series is convergent.3 a d a d a d a d a ad d a ad d ad d d a d ddr dd r ++ =++ + + = + + +=  =− −+==−+ = (ii) Even-numbered terms of GP: 35 2 2 2, , ,...3 3 3a a a           2 2 63 521 3 a aS = = −

Y5 Topical Revision Package 1 5. Sequences and Series (solutions) 7 ( ) ( ) 0 6 2 1 052 61 2 1 05 2 10 61since 0, 2 0 5 2 10 Using GC, 61when 22, 2 0.1 05 2 10 61when 23, 2 1.1 05 2 10 least 23 nSA an a n d an a n a nna nnn nnn n + + + −   −  + + −    − + −   −= + − =  −= + − =−  = 8 DHS PROMOS 2010/Q8 (i) Method 1 (considering the sides) 1st term = 4 cm, common diff = 2 cm Total perimeter S30  304 2(4) (30 1)(2)2 3960 cm =  + − = Method 2 (considering the perimeter) 1st term, a = 16 cm, common diff , d = 8 cm Total perimeter S30  30 2(16) (30 1)(8)2 3960 cm = + − = (ii) (iii) 10000nS   4 2(4) ( 1)(2) 100002 n n + −    2 2(16) ( 1)(8) 100002 4 12 10000 0 n n nn  + −  + −  51.5 48.5n−   Thus largest n = 48. For largest square length, 48 4 (48 1)(2) 98 cm T = + − =

Y5 Top