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