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Mathematics SM0013 Topic 6 : Sequences and Series –Tutorial__________________________________________________________________________________________

CHAPTER 6: SEQUENCE AND SERIES Solution 1. (a) 2. (a) (b) (c)

 2r  1 r 1 4

8

(b)

 (6  r) 3 (c) r 1

19

 2r  3   (1) r 1  r  6    r 1
14

(d)

r r 1

n

r
2

 k  1 k 1 5

=(1+1)+(2+1)+(3+1)+(4+1)=14

 1   1  = 1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1 =8 i i 1

k k 1 8

1

1 1 1 1 1 17 =     2 1 2 3 4 5 60

3. (a) 2,5,8,11,…. Where a = 2 and d = 5-2 = 3 T5= a+4d = 2+4(3) = 14 Tn= a+(n-1)d = 2+(n-1)(3) = 3n-1 Therefore T100= 3(100)-1=299

(b) Let a=2, d=s T5= a+4d = 2+4(s) = 2+4s Tn= a+(n-1)d = 2+(n-1)(s) T100= 2+99s

4. (a) T10 

55 7 , T2  2 2 55 7 a  9d  a  d  ……………(ii) ………(i) and 2 2 (i)-(ii): 8d=24 We obtain d=3 and substitute in (i) 1 we obtain a= 2

(b) 1,4,7,….,88 Let a=1 and d = 4-1=3 88= a+(n-1)d = 1+(n-1)(3) Therefore n=30 (c) 4,x,y,z, 10 Let a=4 and a+4d =10 3 1 1  x  a  d  5 , y  a  2d  7, z  a  3d  8 d= 2 2 2 5. (a) 1+5+9+……+401 Let a=1, d=4 401=1+(n-1)4 n=101

1

Mathematics SM0013 Topic 6 : Sequences and Series –Tutorial__________________________________________________________________________________________

S101 
10

101 1  401 =20301 2
10 10 k 0 k 0

(b)

 3  0.25k  = 31  0.25 k k 0

=3(11)+0.25[0+1+2+….+10] 11  = 33  0.25 0  10  2  =46.75 (c)

 1  2n= 1 - 2 n n 0 n 0 n 0

20

20

20

=[ 1+1+….+1]-2[0+1+2+….+20]  21  = 21  2 0  20 2  = -399 6. (a) 200+202+204+….+1998 a=200 , d=2 1998=200+ (n-1)2 n=900 900 200  1998=989100 S 900  2 n (b) S n  4n  20 2 (i) Sn=2n(n+5) Sn-1=2(n-1)[(n-1)+5] =2(n-1)(n+4) (ii) T1 = S1 = 2(1)(1+5)=12 Since T1+T2 = S2 12+T2=2(2)(2+5) T2=16 d = T2-T1 =16-12 =4 7. (a) 3+6+12+…6 terms Let a=3, r =2 3(26  1) =189 S6  2 1 1 1 (b) 1    ...., 20