1

Example 13

The first two terms of a geometric progression are 3 and –2. Find the least value of n for which the difference between sum of the first n terms and sum to infinity is within 2% of the sum to infinity.

3a3

2rand nSS S

100

2

)1

(100

2

1

)1(

1 r

a

r

ra

r

a n

2

32

1

3

100

2

32

1

)1(3

32

1

3 nr

100

2

3

2

n

n

3

2

:Noten

3

2

n

3

2

5

9

100

21

5

9

5

9 nr

5

9

100

2

5

9 nr

3

02.032

n

02.0lg3

2lg

n

65.9n

Hence the least value of n is 10. (ans)

 

Since

4

Example 14

The sum of the first n terms of a series is

Find the fifth term. Prove that the series is arithmetic and state its common difference.

.2 2 nn

SolutionGiven .nnSn 22

Therefore, 455 SSU

]4)4(2[]5)5(2[ 22 19

5

To prove that the series is an A.P., prove

tconsUU nn tan1

Given ,nnSn 22

112 21 nnSn

1 nnn SSU

)]1()1(2[)2( 22 nnnn

)1242(2 22 nnnnn14 n

1141 nUn

6

)tan(,45414

]1)1(4[)14(1

tconsann

nnUU nn

Thus, the series is arithmetic with common difference 4.

7

Example 15

The sum of the first n terms of a series is

given by . By finding an expression

for the nth term of the series, show that this is a geometric series, and state the value of the first term and the common ratio.

1

1

3

26

n

n

nS

1 nnn SSU

Solution

]3

26[]

3

26[

1)1(

1)1(

1

1

n

n

n

n

8

21

1

3

2

3

2

n

n

n

n

9

3

2

3

3

22n

n

n

n

n

n

n

n

3

29

3

26

n

nU

3

23

To show G.P., show

tconsU

U

n

n tan1

11

3

23

3

23

n

n

n

nU

U

)tan(3

2tconsa

9

Therefore, the series is geometric with first term

3

231U

2

and common ratio 3

2

10

Example 16

The sum of the first 100 terms of an Arithmetic Progression is 10, 000; the first, second and fifth terms of this progression are three consecutive terms of a Geometric Progression. Find the first term, a , and the non-zero common difference, d, of the A.P.Solution 1000011002

2

100100 daS

200992 da ----(1)

a, a+d, a+ 4d are three consecutive terms of a Geometric Progression

11

da

da

a

daratiocommon

4

daada 42

adadada 42 222

add 22 ad 2 ----(2)

Substituting (2) into (1), 200)2(992 aa

1a

From (2), 2d

12

ru rr 23 1

Example 17

The rth term of a series is . Find the

sum of the first n terms.

rr 23 1

Solution

The sum of the first n terms is:

nn uuuuS ...321

)(23...)3(23)2(23)1(23 1131211 nS nn

Given

13

nnn

12

213

131

GP: a = 1, r = 3, n terms

AP: a = 1, d = 1, n terms

)(2...)3(2)2(2)1(23...333 1131211 nS nn

nS nn ...32123...931 1

)1()13(2

1nnn

222132

1nnn

14

Example 18Each time that a ball falls vertically on to a horizontal floor it rebounds to three-quarters of the height from which it fell. It is initially dropped from a point 4 m above the floor. Find, and simplify, an expression for the total distance the ball travels until it is about to touch the floor for the (n+1)th time. Hence find the number of times the ball has bounced when it has traveled 24 m and also the total distance it travels before coming to rest. (The dimensions of the ball are to be ignored.)

15

4m 4

4

3

1st 2nd

44

3 2

3rd nth (n+1)th

44

3 n

Total distance (in metres) that the ball travels

4

4

4

32

4

4

32

2 4

4

32

n

n

4

3.......

4

3

4

3

4

384

32

16

4

31

4

31

4

3

84

n

r

raS

n

n

1

1

callRe

n

4

31244

n

4

32428

17

Hence find the number of times the ball has bounced when it has traveled 24 m and also the total distance it travels before coming to rest

When the ball has traveled 24 m,

244

32428

n

Solution n

4

3244

6

1

4

3

n

6

1lg

4

3lg

n

04

3lg:

Note

Let the number of times the ball has bounced be n.

18

4

3lg

6

1lg

n

23.6n

Since n is an integer, least n=7. Therefore, the ball has bounced 7 times when it has traveled 24 m.

04

3,

nnAs

284

32428 ,

ntherefore

The ball travels 28 m before coming to rest.

For the ball to come to rest, n