736 ANSWERS
5au 1 =3, u 2 =13, u 3 =34, u 4 =70, u 5 = 125,
u 6 = 203,u 7 = 308
bun=^23 n^3 +^32 n^2 +^56 norun=n(n+ 1)(4n+5)
6
c u 50 = 87 125
6au 1 =1,u 2 =5,u 3 =14,u 4 =30,u 5 =55,u 6 =91,
u 7 = 140
bun=^13 n^3 +^12 n^2 +^16 norun=
n(n+ 1)(2n+1)
6
c 338 350
7au 3 =20,u 4 =40,u 5 =70,u 6 = 112
bun=^13 n^3 +n^2 +^23 norun=n(n+ 1)(n+2)
3
c 1360
REVIEW SET 26A
1aStart with 6 and then add 4 successively to get further terms:
u 6 =26,u 7 =30.
bStart with 810 and then multiply each successive term by^13 :
u 5 =10,u 6 =3^13.
2a2 , 5 , 8 , 11 , 14 , ......b4 , 10 , 16 , 22 , 28 , ......
3au 1 =5, u 2 =11,u 3 =17,u 4 =23
bu 1 =4, u 2 =12,u 3 =22,u 4 =34
4aun=4n biun=4n¡ 3 ii un=
1
4 n¡ 1
ciu 20 =77 iiu 20 = 791
5au 1 =18,u 2 =12,u 3 =8,u 4 =5^13
bu 1 =5, u 2 =¡ 10 ,u 3 =20, u 4 =¡ 40
6aun=3£ 4 n¡^1 bun=88£¡
¡^12¢n¡ 17aun=7n¡ 2 bun=n^2 +4n¡ 6
8au 3 =9, u 4 =13,u 5 =17 b un=4n¡ 3
c u 50 = 197
9aun=n^3 +3n^2 +2n
bu 1 =6=1£ 2 £ 3
u 2 =24=2£ 3 £ 4
u 3 =60=3£ 4 £ 5
u 4 = 120 = 4£ 5 £ 6
suggesting that un=n(n+ 1)(n+2)
10 a u 4 =11,u 5 =16
bun=^12 n^2 +^12 n+1orun=
n^2 +n+2
2
c u 10 =56pieces
REVIEW SET 26B
1aStart with 17 and then subtract 5 successively to get further
terms:u 5 =¡ 3 , u 6 =¡ 8.
bStart with¡ 2 and then multiply each successive term by¡ 2 :
u 5 =¡ 32 , u 6 =64.
2a4 , 8 , 12 , 16 , 20 , ......b3 , 8 , 15 , 24 , 35 , ......
3au 1 =1, u 2 =¡ 3 ,u 3 =¡ 7 , u 4 =¡ 11b u 1 =0, u 2 =4, u 3 =10, u 4 =18
4aun=6n biun=6n+4 ii un=^6 n¡^1
6 n+1
ciu 15 =94 iiu 15 =^8991
5aun=n^2 +1 b un=^1
n^3
6a 5 , 15 , 45 , 135 , ...... b ¡ 12 , 3 ,¡^34 , 163 , ......
7aun=4£(¡3)n¡^1 b un= 224£(^14 )n¡^1
8aun=52¡ 9 n b un=n^2 +5n¡ 2
9au 3 =24,u 4 =40,u 5 =60,u 6 =84
b un=2n^2 +2n orun=2n(n+1) c u 10 = 220
10 c un= 241 n^4 + 121 n^3 ¡ 241 n^2 ¡ 121 n
EXERCISE 27A.1
1ax=27 bx=30 c x=18 dx=30
e a=40,b=50 fa=60,b=40
ga=55,b=55 hm=56 i n=49
3 1 cm 440 cm
5a iAbPO=ao iiBbPO=bo
b AbPO=a+b
) a+a+b+b= 180o fangle sum of a¢getc.
c The ‘angle in a semi-circle’ theorem.
6a¢OAB is isosceles as OA=OB fequal radiig
b X is the midpoint of chord AB, ObAX=ObBX,
AbOX=BbOX.
7bThe triangle are congruentfRHSgas:
² OA=AB fequal radiig
² ObAP=ObBP=90o ftangent propertyg
² OP is common
c Consequences are: ² AP=BP
² OP bisects AbPB and AbOB
EXERCISE 27A.2
1ax=64 bx=70 c x=45 dx=66
e x=94 fx=25
2ax=46 bx=30,y=30 c a=50,b=40
d a=55,c=70 ea=80,b= 200
f x=75, y= 118 gx=42 hx=25 i x=25
3 x=70
4aequal radii
bia iib iii 2 a iv 2 b v a+b vi 2 a+2b
c The angle at the centre of a circle is twice the angle at the
circle subtended by the same arc.
5a 2 ® b® c AbDB=AbCB
EXERCISE 27B.1
1ax= 107 fopp. angles of cyclic quad.g
b x=60fopp. angles of cyclic quad.g
c x=70fopp. angles of cyclic quad.g
d x=81fexterior angles of cyclic quad.g
e x=90fexterior angles of cyclic quad.g
f x= 125 fexterior angles of cyclic quad.g
2ax= 110,y= 100 b x=40
c x=65, y= 115 d x=80
e x=45, y=90 f x= 105,y= 150
3 BbAD=95o,AbDC=65o,DbCB=85o,AbBC= 115o, , .....
, , ....., , .....
, , .....IB MYP_3 ANS
cyan magenta yellow black(^05255075950525507595)
100 100
(^05255075950525507595)
100 100
Y:\HAESE\IGCSE01\IG01_an\736IB_IGC1_an.CDR Thursday, 20 November 2008 11:47:46 AM PETER