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Sequence and series Questions

 

1.  The six exterior angles of a hexagon form an arithmetic progression. if the smallest angle is 150, find the size of the biggest angle of the hexagon (3 mks)

2.  An arithmetic progression whose first term is 2 and whose nth term is 32 has the sum of its first n terms equal to 557. find n. (3mks)

3.  Every time an insect jumps forward the distance covered is half of the previous jump. If the insect initially jumped 8.4 cm, calculate

(i) To the nearest two decimal places the distance of the sixth jump (1 mk)

(ii) The total distance covered after the sixth jump (2mks)

4.  (a) An arithmetic progression is such that the first term is -5, the last is 135 and the sum of

progression is 975.

 Calculate

 (i) The number of terms in the series. (7 marks)

 (ii) The common difference of the progression (2 marks)

(b) The sum of the first three terms of a geometric progression is 27 and the first term is 36.

Determine the common ratio and the value of the fourth term.  (4 marks)

5.  The sum of the second and fourteenth terms of an arithmetic progression is 15 and sum of the fifth and sixth terms is 25. Find the first term and the common difference. (4 mks)

6.  The area covered by Mau forest is 40,000 km2 at present. If the human encroachment rate is

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estimated to be 2 % every 10 years. Calculate the area of the forest encroached in 30 years.

 

7.  Three consecutive terms of geometric progression are 3 2x + 1, 9x and 81 respectively. Calculate:

 (a) The value of x  

 (b) Find the common ratio

 (c) Calculate the sum of the first 10 terms of this series

 (d) Given that the fifth and the seventh terms of this G.P forms the first two consecutive

  terms of arithmetic sequence, calculate the sum of the first 20 terms of this sequence  

 

8.  How many terms of the sequence -12 + -10 + -8 …..should be added to give a sum of 338?

 

9.  An arithmetic progression whose first term is 2 and whose nth term is 32 has the sum

of its n terms equal to 357. Find n  

 

 

 

10.

 

 

 

 

 

 

 

 

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 In the figure OACB is parallelogram in which M is the mid- point of AC and OM produced

  meets BC also produced at X.

 Given OA = a and OB = b

 a) Express OC in terms of a and b  

 b) Find the values of r and s such that OX = rOM and CX = sBC

 c) Hence determine the ratio BC:BX  

11.  For the series 29 + 23 +….. + (-91), find;

 (a) The number of terms in the above series  

 (b) The sum of the series

12. (a) Given that 5, a, b, and 7 are in arithmetical progression, find the value of a and b  

(b) If 5, P, Q , 135/8 are in geometrical progression. Find the value of P and Q

(c) Prove that the sum of the first 12 terms of the first series in (a) is approximately

equal to the sum of the first 6 terms of the second series (b) above

13.  An aeroplane flew East for 640km then turned and flew on a bearing of 050o. After 2.5hrs

flying at 324km/hr, it was necessary to fly to the original point because of technical hitch.

How much shorter is it going to cover flying straight to the starting point than retracing

its former route?  

14.  A ball falls vertically from a height of 15m. Each time it bounces back to 50% of the

height achieved on the previous bounce. Find the distance covered after 6 such bounces  

15.  Find the sum of the first 51 terms of the series:-

 -22, -19, -16……………………………..

16.  Olunga saves shs.100 on his son’s first birthday. He saves shs.200 on the second birthday and

  Shs.400 on the third birthday and so on doubling the amount on every birthday. How much

  will he be saving on the boy’s 10th birthday.

17.  A self-help group intended to purchase a dry cleaning machine worth shs.720,000.

The members  were required to contribute equal amounts to pay for the machine.

The group recruited 20 more members consequently, each member paid shs.3000 less

that what he would have contributed.

 (a) find the original number of members  

 (b) find the amount required from each member to contribute after the recruitment

18.  Find the number of terms in the following sequence

 8, 4, 2, ½ …….., 1/512

19.  An arithmetic progression has the first term a and the common difference d

 a) Write down the third, ninth and twenty fifth terms of the progression in terms of a and d

 b) The arithmetic progression above is increasing and that the third, ninth and twenty fifth

terms form the first three consecutive terms of a geometric progression. The sum of the

seventh and twice the sixth terms of the arithmetic progression is 78.

Calculate:

i) The first term and common difference of the arithmetic progression

ii) The sum of the first nine terms of the arithmetic progression

20.  The difference between the fourth and the seventh terms of an increasing arithmetic progression

 

 

 

Sequence and series Answers

1

Image From EcoleBooks.com

Image From EcoleBooks.com

Image From EcoleBooks.com

 

M1

M1

 

A1

Alternative

Image From EcoleBooks.com

 

Image From EcoleBooks.com

  

3

 

2.

a = 2

nth = 32

Sn = 357

Sn = n/2 (2 + 32)

714 = 34n

21 = n

 

M1

 

M1

 

A1

 

Substitution

 

Simplification

  

03

 

 

1.   P 1 + R

Image From EcoleBooks.com
100

 

  = 40,000 1 + 2

Image From EcoleBooks.com 100

 

 = 40, 000 x (1.02) = 42,448.32 km2

  Encrouched area

= 42 448.32 – 40 000 = 2448.32 km2

 

2.  (a) 9x = 81

   32x + 1 9x

 92x = 34(32x + 1)

  34x = 34 + 2x + 1

  34x = 32x + 5

  4x = 2x + 5

  2x = 5

  x = 2.5

 

 (b) Common ratio = 81

 92.5

 = 1

3

Image From EcoleBooks.com  (c) a = 3(2 x 2.5 + 1)

= 36

= 729

  S10 = 729 1 – 1
10

3

  1 – 2

3

= 1093.5 x 0.99998 = 1093.5

 

(d) 5th term = 729 x (1/3)4

= 9

 7th term = 729 x (1/3)6


= 1

  a = 9 d = 1 -9 = -8

S20 = 20 2 x 9 + (20 – 1) (-8)

2

 

  = 10 (18 – 152) = -1340

 

 

3. – 12 + – 10 + -8 + ……………………..

  a = -12 d = z

 

Sn = n 2a + (n – 1)d

2

 

 338 = n 2(-12) + (n- 1)2

2

 

 676 = n – 24 + 2n – 2

 

  2n226n – 676 = 0

  2 2 2 2

 

n2 – 13n – 338 = 0

(n – 26) (n + 13) = 0

  n = 26 or n = -13 reject


∴ n = 26 terms


 

3. – 12 + – 10 + -8 + ……………………..

  a = -12 d = z

 

Sn = n 2a + (n – 1)d

2

 

 338 = n 2(-12) + (n- 1)2

2

 

 676 = n – 24 + 2n – 2

 

  2n226n – 676 = 0

  2 2 2 2

n2 – 13n – 338 = 0

(n – 26) (n + 13) = 0

  n = 26 or n = -13 reject


∴ n = 26 terms

 

4.  32 = 2 + (n-l)d…….(i)

357 = n 2.2 + (n-l)d …… (ii)

2

 N 4 + (n-l) d = 714

 2 + (n-l)d = 32

 N(4 + nd – d) = 714

 -d + nd = 30

 4n + n2d – d = 744

 nd – d = 30

 d(n-l) = 30

 

5.  a) OC = OB + BC = a + b

 

b) OM = OA + AM = a + ½ b

  Given OX = rOM

= r(a + ½ b)

From OBX

Ox = OB + BX

= OB + BC + CX

= b + a + sa

= (1+s) a +b

 r(a + ½ b) = (1 + s) a + b

Comparing coefficients of a and b

  r = 1 + S

and ½ r = 1 r = 2

Substitute for r = 2  2 = 1 + s  s = 1

 

 

c) Now BX = BC + Cx

  = a + a = 2a

 

 BC:BX = 1:2

6.  (a) -91 = 29 + (n-1) x -6

  -120 = -6n + 6

  6n = 126

  n = 21

 

(b) S21 = 21/2 [(2 x 2a) + (20 x -6)]

= 21 x -62

2

  = -651

7. d = p-5…..(i)

d = q – p ………(ii)

0 = 2p -q – 5

0 = 7 – 2q + p

– p + 2q = 7

2p – q = 5

 

-2p + 4q = 14

2p – q = 5

3q = 19

q = 19/2

p=2q – 7 38/3 – 7

p=17/8

S = n [2a + (n-1)d}

2

= 12/2 (10+11 x 2/3)

= 6 (10+22/3) = 104

Sn = a(rn-1) = S(1.5-6)

r-1 1.5 – 1

= 5 x (1.5 – 1) = 103.90

 0.65 = 10.4  

 

8.  a + a + d = 10………………(i)

 

10 2a + 9d = 210……………..(ii)

 2

2a+ d =10

2a + 9d = 42

8d = 32

d= 4

T1 = 3 +6(4)

= 3 + 24

= 27

 

9.  S6 = 15 (1-0.56)

1- 0.5

= 29.5314metres

 

 

 

 

10.  Sn = n 2a + (n-1)L

2

S51 = 51 (2x -22) + (51 -1)3

2

   = 2703

11.  100 + 200 + 400 + 800 + 1600 + 3200 + 6400 + 12800 + 25600 + 51200

200 = 400 = 800

 100 200 400

= 51200 99600 108200 110,600

  25600
6400
1600
700

 76800 105,000 109,800 111,300

  12800
3,200
800

  99,600 108,200 110,600

= 111300

 

 

12..  a) Let n be the initial members

Each to contribute 720000

n

New membership n + 20

Contributions: 720000

  n + 20

720000 – 720000 = 3000

  n n + 20

 720000(n + 20) – 720000n= 3000n(n + 20)

4800 = n(n + 20)

  n2 + 20 – 4800 = 0

  n2 + 80n – 60n – 4800 = 0

  n(n + 80) – 60(n + 80)= 0

(n-60) (n + 80) = 0

n = 60

Original members = 60

 

b) Contributions required before recruitment

 = 720000

60 = 120000

After requirement = 720000

 

13.  nth term is arn-1

a = 8, r = ½

 

nth term = 1/512

8( ½ )n-1 = 1/512

 

8( ½ )n-1 = 2-9

( ½ )n-1 = 2-9 ÷ 23

( ½ )n-1 = 2-12 = ( ½ )12

n-1 = 12

  n = 13

 

14. 3rd a + 2d

9th a + 8d

25th a + 24d

(i)   a + 2d = a + 8d

a + 8d a = 24d

 (a +2d) (a +2d) = (a + 8d) (a +8d)

 a2 + 26da + 48d2 = a2 + 16da + 64d2

 10da = 16d2

 10d 10d

 a = 1.6d………….(i)

 (a + 6b) + 2(a+5d) = 78

 3a + 16d = 78…………………(ii)

 But a = 1.6d

 (3×1.6d) +16d = 78

 4.8d + 16d = 78

 4.8d + 16d = 78

 20.8 = 78

 20.8 20.8

 Common distance d= 3.75

 a=1.6 x 3.75

 first term a = 6

 

(ii)   Sn = n/2 (2a + (n-1)d)

Sa = 9/2 ((2×6) + (9-1)3.75)

= 9/2 (12 + 30)

9/2 x 42= 189

 

15.  T4 = a + 3d

T7 = a + bd

(a + 6d) – (a + 3d) = 12

3d = 12

d = 4

But a = 9

S5 = 5 2(9) + 4 (4)

  2

= 5 18 + 16

  2

= 5 x 34

  2

  = 85


 




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