1

Example 1:

Find the 10th term and the nth term for the sequence 7, 10, 13, … .

Solution:7a 3d

U10 = 34

31107

Un= 43

317

nn

2

Example 2

Find the three numbers in an arithmetic progression whose sum is 24 and whose product is 480.

Solution

Let the three numbers be (a-d), a, (a+d) where d is the common difference.

(a-d) +a + (a+d) = 243a = 24

a = 8

(a-d)(a)(a+d)=480Substituting a =8, we get

2d

3

When d = 2, required numbers are 6, 8, 10When d = -2 , required numbers are 10, 8, 6

4

Example 3

The sum of the first eighteen terms of an arithmetic series is -45 and the eighteenth term is also -45.

Find the common difference and the sum of the first hundred terms.

Solution

])118(2[2

184518 daS

4515318 da --------- (1)

122

dnan

Sn

4518 U

5

4518 U

45118 da4517 da ------------ (2)

Solving equation 1 and 2, we get

5

40

d

a

6

Example 4

Find the sum of the positive integers which are less than 500 and are not multiples of 11.

Solution

Required sum=Sum of integers from 1 to 499(SI)

Sum of multiples of 11(SII)

499............4321 IS

49912

499

124750

7

495.......44332211 IIS 45........432111

451

2

4511

11385

lan

Sn 2

Required sum = SI-SII

=124750-11385

=113365

8

Example 5

The sum of the first 10 terms of an A.P. is 145 and the sum of the next 6 terms is 231

Find (i) the 31st term , and

(ii) the least number of terms required for the sum to exceed 2000.

daU 3031

14510 S

231161514131211 UUUUUU10S 145

16S da 1522

16

9

Solution

2311451522

1616 daS

3761528 da47152 da (1)

14510 S

145922

10 da

(2)2992 da

10

(1)(2): 186 d3d

From (2):

12

3929

a

9133013031 daU

(ii)Find the least number of terms required for the sum to exceed 2000.

Let n be the least number of terms required for 2000nS

11

2000122

dnan

200031122

nn

400013 nn

040003 2 nn

Consider 040003 2 nn

3.36or 7.366

40003411

n

12

3.36 7.36++

7.36or 3.36 nn

Since n must be a positive integer, n > 36.7

Hence, the least number of terms required is 37.

040003 2 nnFor

13

Example 6

Given that the fifth term of a geometric

progression is and the third term is 9.

Find the first term and the common ratio if all the terms in the G.P. are positive.

4

81

4

814 ar

92 ar

(1)

(2)

14

(1)(2):9

1

4

812

4

ar

ar

4

92 r

Since all the terms in the G.P. are positive, r > 0

2

3r

From (2): 92

3 2

a

4a

15

Example 7

Three consecutive terms of a geometric

progression are and 81. Find the value of x. If 81 is the fifth term of the geometric progression, find the seventh term.

13 ,3 xx

x

x

r3

3 1

13

81

x

1

41

3

3

3

3

xx

x

16

x 333x31

2xGiven 81 is the fifth term, find the seventh term:

U5= ar4 = 81

33

3

3

813

4

1 x

r

813 4 a

13

814a

U7 = ar6 729)3(1 6

17

Example 8

Find the sum of the first eight terms of the series

Soln:.........

9

8

3

423

G.P. with ,3a3

2r

3

2 ,

3

2 ,

3

2

3

4

2

3

1

2 U

U

U

U

U

U

r

raS

1

1 8

8

3

21

3

213

8

18

3

21

3

213

8

36561

25613

729

6305

19

Example 9

A geometric series has first term 1 and the common ratio r, where r 1, is positive. The sum of the first five terms is twice the sum of the terms from the 6th to 15th inclusive. Prove that

).13(2

15 r

1a

5S

1576 ..... UUU 157654321 ........ UUUUUUUU

515 SS 54321 UUUUU

20

SolutionGiven 5S 5152 SS

155 23 SS

1

)1(2

1

)1(3

155

r

ra

r

ra

Since and a = 1, we have 1r

)1(2)1(3 155 rr

0132 515 rr

013)(2 535 rr

21

Let .5rx So the equation becomes 0132 3 xx

Let f(x) = 132 3 xxSince f(1) = 2 – 3 + 1 = 0

(x –1) is a factor of f(x).

132 3 xx )12)(1( 2 kxxx

Comparing coefficient of x:

2

13

k

k

132 3 xx )122)(1( 2 xxx

22

Hence, 0)122)(1( 2 xxx

01221 2 xxorx

)2(2

)1)(2(442 x

)13(2

1)13(

2

1 or

4

322

4

1221 55

rorr

Since 1r

1r

and r > 0, )13(2

15 r

(r<0)

23

SSum to infinity, S (or ) = r

a

1

The sum to infinity exists (the series converges or series is convergent) provided 1r

Example 10

Determine whether the series given below converge. If they do, give their sum to infinity.

........16842 (a)G.P. with >12r

Does not converge

24

(b) ...........64

1

16

1

4

11 G.P. with

4

1r

Series converges4

1r <1

r

aS

1

4

11

1

5

4

25

(c) ....................4

27

2

93 G.P. with

2

3r

2

3r >1

Does not converge

26

A geometric series has first term a and the common ratio .

Show that the sum to infinity of the geometric progression is .

Example 11

2

1

)22( a

Solutionr

aS

1

2

11

a

12

2

a

12

12

12

)22(

a )22( a (shown)

27

Example 12

A geometric series has first term a and common ratio r. S is the sum to infinity of the series, T is the sum to infinity of the even-numbered terms (i.e. ) of the series. Given that S is four times the value of T, find the value of r.

....642 UUU

r

aS

1arT 3ar .........5 ar

1

ar2r

Given S = 4T, we have r

a

1 214

r

ar

common ratio r 2

28

)1)(1(

4

1

1

rr

r

r

r

r

1

41

rr 41

3

1r