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## SEQUENCE AND SERIES

SEQUENCE

Is a set of numbers written in a definite order such that there is a rule by which the terms are obtained.

Or

Or

Is a set of number with a simple pattern.

Example

1. A set of even numbers

• 2, 4, 6, 8, 10 ……

2. A set of odd numbers

• 1, 3, 5, 7, 9, 11….

Knowing the pattern the next number from the previous can be obtained.

Example

1. Find the next term from the sequence

• 2, 7, 12, 17, 22, 27, 32

The next term is 37.

2. Given the sequence

• 2, 4, 6, 8, 10, 12………

What is

i)The first term =2

ii) The 3rd term =6

iii)The 5th term =10

iv)The nth term [ the general formula]

2=2×1

4=2×2

6=2×3

8=2×4

10=2×5

12=2×6

Nth =2xn

Therefore nth term =2n

Find the 100th term, general formula =2n

100th term means n=100

100th term =2×100

100th term =200

3. Find the nth term

,

Solution

,

4. Given the general term 3[2n]

a) Find the first 5terms

b) Find the sum of the first 3 terms

Solution

3[2n] when;

n=1, 3 [21] =6

n=2, 3 [22] =12

n=3, 3 [23] =24

n=4, 3 [24] =48

n=5, 3 [25] =96

First 5 terms = 6, 12, 24, 45, 96

Sum of the three terms

Sum of the first three = 6+12+24

Sum of the first three = 42

Exercise 1.

1. Find the nth term of the sequence 1, 3, 5, 7…..

2. Find the nth term of the sequence 3, 6, 9, 12……

3. The nth term of a sequence is given by 2n+1 write down the 10th term.

4. The nth term of a certain sequence is 2n-1 find the sum of the first three terms.

5. If the general term of a certain sequence is

Find the first four terms increasing or

decreasing in magnitudes

Solution

2. Find the nth term of the sequence 3, 6, 9, 12……

3. The nth term of a sequence is given by 2n+1 write down the 10th term.

4. The nth term of a certain sequence is 2n-1 find the sum of the first three terms.

5. If the general term of a certain sequence is

Find the first four terms increasing or

decreasing in magnitudes

Solution

1. 1, 3, 5, 7…nth

From the sequence the difference between the consecutive terms is 2 thus

nth =2+n

2. 3, 6, 9, 12……..nth

2. 3, 6, 9, 12……..nth

The difference between the consecutive terms is 3 thus

nth =3n

3. 10th = 2[10+1]

10th =20+1

10th =21

4. When

n=1, 21-1

n=2, 22-1

n=3, 23-1

Sum of the first three terms =1, 2, 4

Sum = 1+2+4

Sum = 7

5. When

n = 1, = 1

n = 2, =

n = 3, = 1

n = 4, =

The first four terms are

SERIES

Defination: When the terms of a sequence are considered as the sum, the expression obtained is called a serier or a propagation

Example

(a). 1+2+3+4+5+……….

(b). 2+4+6+8+10+………+100.

(c). -3-6-9-…….

The above expression represent a series. There are two types of series

1.Finite series

Finite series is the series which ends after a finite number of terms

e.g. 2+4+6+8+………….+100.

-3-6-9-12-………….-27.

2. infinite Series

Is a series which does not have an end.

e.g. 1+2+3+4+5+6+7+8…………..

1-1+1-1+1-1+1-1+1…………..

Exercise 5.2.1

1. Find the series of a certain sequence having 2(-1)n as the general term

2. Find the sum of the first ten terms of the series -4-1+2+…….

3. The first term of a certain series is k. The second term is 2k and the third is 3k. Find

(a). The nth term

(b). The sum of the first five terms

Exercise 5.2.1 Solution

1. n=1, 2(-1)1 = -2

n=2, 2(-1)2 = 2

n=3, 2(-1)3 = -2

n=4, 2(-1)4 = 2

n=5, 2(-1)5 = -2

The series is -2+2-2+2-2+2-2+2-2+2+………………..+2(-1)n

2. The sum of the first n terms of the series -4-1+2+5+8+11+14+17+20+23

= 95

3. (a) k + 2k + 3k + 4k +………. + nk

The nth term of the series is nk

(b) k + 2k + 3k + 4k + 5k = 15k

... The sum of the first 5 terms = 15k

ARITHMETIC PROGRESSION [A.P]

An arithmetic progression is a series in which each term differ from the preceding by a constant quantity known as the common difference which is denoted by “d“.

Defination: When the terms of a sequence are considered as the sum, the expression obtained is called a serier or a propagation

Example

(a). 1+2+3+4+5+……….

(b). 2+4+6+8+10+………+100.

(c). -3-6-9-…….

The above expression represent a series. There are two types of series

1.Finite series

Finite series is the series which ends after a finite number of terms

e.g. 2+4+6+8+………….+100.

-3-6-9-12-………….-27.

2. infinite Series

Is a series which does not have an end.

e.g. 1+2+3+4+5+6+7+8…………..

1-1+1-1+1-1+1-1+1…………..

Exercise 5.2.1

1. Find the series of a certain sequence having 2(-1)n as the general term

2. Find the sum of the first ten terms of the series -4-1+2+…….

3. The first term of a certain series is k. The second term is 2k and the third is 3k. Find

(a). The nth term

(b). The sum of the first five terms

Exercise 5.2.1 Solution

1. n=1, 2(-1)1 = -2

n=2, 2(-1)2 = 2

n=3, 2(-1)3 = -2

n=4, 2(-1)4 = 2

n=5, 2(-1)5 = -2

The series is -2+2-2+2-2+2-2+2-2+2+………………..+2(-1)n

2. The sum of the first n terms of the series -4-1+2+5+8+11+14+17+20+23

= 95

3. (a) k + 2k + 3k + 4k +………. + nk

The nth term of the series is nk

(b) k + 2k + 3k + 4k + 5k = 15k

... The sum of the first 5 terms = 15k

ARITHMETIC PROGRESSION [A.P]

An arithmetic progression is a series in which each term differ from the preceding by a constant quantity known as the common difference which is denoted by “d“.

For instance 3, 6, 9, 12….. is an arithmetic progression with common difference 3.

The nth term of an arithmetic progression

If n is the number of terms of an arithmetic progression, then the nth term is denoted by An

Therefore An+1= An + d

e.g. First term = A1

Second term = A1+ d = A2

Third term = A2 + d = A3

Consider a series 3+6+9+12+…..

A1 = 3, d= 3

A2 = A1 + d

A3 = A2 + d = A1+d+d = A1+ 2d

A4 = A3 + d = A1+ 2d+ d = A1 + 3d

A5 = A4 + d = A1+ 3d+ d = A1 + 4d

A6 = A5 + d = A1+ 4d+ d = A1 + 5d

An = A[n-1] + d = A1+ (n-1)d

... The general formula for obtaining the nth term of the series is

An = A1+ (n-1)d

The general formula for obtaining the nth term in the sequence is also given by An =A1+[n-1]d

The nth term of an arithmetic progression

If n is the number of terms of an arithmetic progression, then the nth term is denoted by An

Therefore An+1= An + d

e.g. First term = A1

Second term = A1+ d = A2

Third term = A2 + d = A3

Consider a series 3+6+9+12+…..

A1 = 3, d= 3

A2 = A1 + d

A3 = A2 + d = A1+d+d = A1+ 2d

A4 = A3 + d = A1+ 2d+ d = A1 + 3d

A5 = A4 + d = A1+ 3d+ d = A1 + 4d

A6 = A5 + d = A1+ 4d+ d = A1 + 5d

An = A[n-1] + d = A1+ (n-1)d

... The general formula for obtaining the nth term of the series is

An = A1+ (n-1)d

The general formula for obtaining the nth term in the sequence is also given by An =A1+[n-1]d

Question

1. A7 = A2 and A4 =16. Find A1 and d.

Solution

A7= A1+6d

=A1+6d= ( A1+d)

A7= A1+6d

=A1+6d= ( A1+d)

2A1+12d=5A1+5d

3A1=7d

A1= d

A1= d

A 4 =16

A4=A1+3d = 16

d +3d= 16

A4=A1+3d = 16

d +3d= 16

d = 16

d= 3

But A1 +3d= 16

A1 +9=16

A1= 7

A1= 7

Exercise 2

1. The pth term of an A.P is x and the qth term of this is y, find the rth term of the same A.P

2. The fifth term of an A.P is 17 and the third term is 11. Find the 13th term of this A.P.

3. The second term of an A.P is 2 and the 16th term is -4 find the first term.

4. The sixth term of an A.P is 14 and the 9th of the same A.P is 20 find 10th term.

5. The second term of an A.P is 3 times the 6th term. If the common difference is -4 find the 1st term and the nth term

6. The third term of an A.P is 0 and the common difference is -2 find;

(a) The first term

(b) The general term

2. The fifth term of an A.P is 17 and the third term is 11. Find the 13th term of this A.P.

3. The second term of an A.P is 2 and the 16th term is -4 find the first term.

4. The sixth term of an A.P is 14 and the 9th of the same A.P is 20 find 10th term.

5. The second term of an A.P is 3 times the 6th term. If the common difference is -4 find the 1st term and the nth term

6. The third term of an A.P is 0 and the common difference is -2 find;

(a) The first term

(b) The general term

7. Find the 54th term of an A.P 100, 97, 94

8. If 4, x, y and 20 are in A.P find x and y

9. Find the 40th term of an A.P 4, 7, 10……..

10.What is the nth term of an A.P 4, 9, 14

8. If 4, x, y and 20 are in A.P find x and y

9. Find the 40th term of an A.P 4, 7, 10……..

10.What is the nth term of an A.P 4, 9, 14

11.The 5th term of an A.P is 40 and the seventh term of the same A.P. is 20 find the

a) The common difference

b) The nth term

12. The 2nd term of an A.P is 7 and the 7th term is 10 find the first term and the common difference

b) The nth term

12. The 2nd term of an A.P is 7 and the 7th term is 10 find the first term and the common difference

Exercise 2 Solution

1. d = y-x

rth = A1+[n-1]d

rth = A1+[n-1]d

= A1+nd-d

= A1+n[y-x]-[y-x]

= A1+[ny-nx-[y-x]

=A1+[ny-y]-[nx-x]

A1+g[n-1]-x[n-1]

rth=A1+[y-x][n-1]

2.

A5=17

A3=11

A13=?

A13=?

A5 =A1+4d = 17

A3=A1+2d =17

A1+2d=11

A3=A1+2d =17

A1+2d=11

Solve the simultaneous equation by using equilibrium method.

A1+4d=17

A1+2d=11

=

A1+2d=11

=

d = 3

A1+4[3]=17

A1+12=17-12

A1=5

A1+12=17-12

A1=5

A13=A1+12d

A13=5+12[3]

A13=5+36

A13 =41

A13=5+12[3]

A13=5+36

A13 =41

3.

A2=2

A16=-4

A1=?

A16=-4

A1=?

A2=A1+d

A1+d=2

A1+d=2

A16=A1+15d

A1 +15d= -4

A1 +15d= -4

Solve the two simultaneously equations by using the elimination method

A1 + d=2

A1 +15d=-4

d=

From the 1st equation

A1 +

A1 = 2+

A1 =

4.

A6= 14

A9=20

A6= A1+5d = 14

A9 = A1+8d=20

Solve the simultaneous equation by using the elimination method

A1+5d= 14

A1+8d=20 ( difference between two equations)

Then d = 2

From 1st equation

A1 +5[2] = 14

A1 +10= 14-10

A1=4

A1 +10= 14-10

A1=4

A10= A1+9d

A10=4+9[2]

A 10=4+18

A10=22

A10=4+9[2]

A 10=4+18

A10=22

An =A1+[n-1]d

An =4[n-1]2

An =4+2n-2

An= 2n+4-2

An=2n+2

An =4[n-1]2

An =4+2n-2

An= 2n+4-2

An=2n+2

5.

A2= 3 x Ab

D= -4

A1=?

An=?

D= -4

A1=?

An=?

Az = 3 x A6

A 1 +d = 3 x A1+5d

A1+d =3A1+15d

2A1+14d=0

d= -4

A 1 +d = 3 x A1+5d

A1+d =3A1+15d

2A1+14d=0

d= -4

2A1+14[-4]=0

2A1+-56= 0

2A1= 56

A1= 28

2A1+-56= 0

2A1= 56

A1= 28

An = A1+[n-1]d

An= 28+[n-1]-4

An=32-4n

An =32-4n

An= 28+[n-1]-4

An=32-4n

An =32-4n

6.

(a) A

3= 0

d= -2

3= 0

d= -2

From the formula

A1+2d= 0

A1+2[-2]= 0

A1-4=0

A1= 4

A1+2[-2]= 0

A1-4=0

A1= 4

The first term is 4.

b) The general term

An =A1+ [n-1] d

A n =4+[n-1]-2

An = 4+ -2n+2

An = 6-2n

A n =4+[n-1]-2

An = 4+ -2n+2

An = 6-2n

The general term is 6-2n.

7.

A54 =?

100, 97, 94 = A1, A2, A3

A54= A1+53d

d= A2-A1= A3-A2

d= 97-100= 94-97

d= -3

100, 97, 94 = A1, A2, A3

A54= A1+53d

d= A2-A1= A3-A2

d= 97-100= 94-97

d= -3

A54= 100+53[-3]

A 54=100 + -159

A54= -59

A 54=100 + -159

A54= -59

8.

4, x, y, 20

A 4= 20

A1 +3d=20 , but A1=4

A 4= 20

A1 +3d=20 , but A1=4

4+3d= 20

3d=16

d = 16/3

3d=16

d = 16/3

A 1, A2, A3, A4

A2 =A1+d

4+

X =

A3 = A1+2d

4+ 2x

A3 =

Hence x= and y=

9.

A40 =?

A1 =4

A2= 7

A 3= 10

A1+39d=?

A1 =4

A2= 7

A 3= 10

A1+39d=?

d= 7-4= 10-7

d= 3

d= 3

A1 +39[3]

A1+117

4+117

A40 =121

A1+117

4+117

A40 =121

10.

A1 =4

A2=9

A3=14

d= 9-4= 14-9

d= 5

A2=9

A3=14

d= 9-4= 14-9

d= 5

An =A1+[n-1]d

An =4+[n-1]5

An =4+5n-5

An =5n-1

An =4+[n-1]5

An =4+5n-5

An =5n-1

The nth term is 5n-1.

11.

a) the common difference

A5= 40

A7= 20

A5= 40

A7= 20

A1+ 4d= 40 ………… (1)

A1+ 6d= 20 ………….(2)

A1+ 6d= 20 ………….(2)

Subtracting equation (2) from equation (1) we obtain

-2d=20

d= -10

d= -10

b) the tenth term

A10= A1+9d,

But A1 +4d=40

A1=80

But A1 +4d=40

A1=80

∴A10=A1+9d

=80-90

A 10=-10

=80-90

A 10=-10

12.

A2= A1+d

A7= A1+6d

A1 +6d = 10

A1+d=7

A7= A1+6d

A1 +6d = 10

A1+d=7

Solving the simultaneous equations by using the elimination method;

-5d= -3

d = 3/5

A1 + = 7

A1 =

d = 3/5

A1 + = 7

A1 =

SUM OF THE FIRST n TERMS OF AN ARITHMETIC PROGRESSION

Consider a series with first term A1, common difference d. If the sum n terms is denoted by Sn, then

Sn = A1 + (A1+ d) + (A1 + 2d)+ …+ (A1 – d) + An

+ Sn = An + (An- d) + (An – 2d)+ …+ (A1 + d) + A1

2Sn = (A1+ An) + (A1+ An) + (A1+ An)+ …+ (A1+ An)+ (A1+ An)

There are n terms of (A1+ An) then

2Sn = n(A1+ An)

Sn = n(A1+ An)

2

... The sum of the first n terms of an A.P with first term A1 and the last termAn is given by

Sn = (A1+ An)

But An =A1+ (n-1) d

Thus, from

Sn = (A1+ An)

Sn = [A1+ A1+ (n-1) d]

Sn = [2A1+ (n-1) d]

... therefore, the sum of the first n term of an A.P with the first A1 and the common difference d in given by

Sn = [2A1+ (n-1) d]

Consider a series with first term A1, common difference d. If the sum n terms is denoted by Sn, then

Sn = A1 + (A1+ d) + (A1 + 2d)+ …+ (A1 – d) + An

+ Sn = An + (An- d) + (An – 2d)+ …+ (A1 + d) + A1

2Sn = (A1+ An) + (A1+ An) + (A1+ An)+ …+ (A1+ An)+ (A1+ An)

There are n terms of (A1+ An) then

2Sn = n(A1+ An)

Sn = n(A1+ An)

2

... The sum of the first n terms of an A.P with first term A1 and the last termAn is given by

Sn = (A1+ An)

But An =A1+ (n-1) d

Thus, from

Sn = (A1+ An)

Sn = [A1+ A1+ (n-1) d]

Sn = [2A1+ (n-1) d]

... therefore, the sum of the first n term of an A.P with the first A1 and the common difference d in given by

Sn = [2A1+ (n-1) d]

Where

n =number of terms

A1= first term

An =last term

d= common difference

A1= first term

An =last term

d= common difference

Example

i) Find sum of the first 5th term where series is 2, 5, 8, 11, 14 first formula

Solution

S5 = [2 + 14]

S5 = 40

S5 = 40

(ii)Sn = [2A1 + [n-1] d]

S5 = [2 x 2 + [5-1] (3)]

S5 = [2 x 2 + [5-1] (3)]

S5= 40

Arithmetic Mean

Arithmetic Mean

If a, m and b are three consecutive terms of an arithmetic. The common difference

d= M – a

Therefore M- a = b – M

2M = a+ b

M= a + b

2

M is called the arithmetic mean of and b

E.g. Find the arithmetic mean of 3 and 27

M = 3+ 27

2

= 30 = 15

2

GEOMETRIC PROGRESSION (G.P)

Defination:

Geometric progression is a series in which each term after the first is obtained by multiplying the preceding term by the fixed number.

The fixed number is called the common ratio denoted by r.

E.g. 1+2+4+8+16+32+…

3+6+12+24+48

The nth term of Geometric Progression

If n is the number of terms of G.P, the nth term is denoted by Gn and common ratio by r. Then G n+1 = rGn for all natural numbers.

G1 = G1

G2 = G1r

G3 = G2r = G1r.r = G1r2

G4 = G3r = G1r2.r = G1r3

G5 = G4r = G1r3.r = G1r4

G6 = G5r = G1r4.r = G1r5

G7 = G6r = G1r5.r = G1r6

G8 = G7r = G1r6.r = G1r7

The nth term is given by

Gn = G1rn-1

Example1: Write down the eighth term of each of the following.

(a). 2+ 4+ 8+…

(b). 12+ 6+ 3+ …

Solution

(a) The first term G1 =2, the common ratio r=2 and n=8, Then from

Gn = G1rn-1

G8 =(2)(2)8-1

G8 = (2). 27

G8 = 256

(b) The first term G1 =12, the common ratio r=1/2 and n=8, Then from

Gn = G1rn-1

G8 =(12)(1/2)8-1

G8 = (12). (1/2)7

G8 = 12/128 = 3/32

Example2: Find the numbers of terms in the following 1+2+4+8+16+…+512.

Solution

The first term G1 =1, the common ratio r=2 and Gn= 512, Then from

Gn = G1rn-1

512 =(1)(2)n-1

512 = 2n-1

512 = 2n.2-1

512 x 2= 2n

1024= 2n

210 = 2n

n = 10

The sum of the first n terms of a geometrical progression.

Let the sum of first n terms of a G.P be denote by Sn

Sn = G1+ G2+ G3+ G4+…+ Gn

Since the common ratio is r

From

G2 = G1r

G3 = G1r2

G4 = G1r3

…Gn = G1rn-1

Sn = G1+ G1r+ G1r2+ G1r3+…+ Gnrn-1

Multiplying by common ratio r both sides we have

rSn

(a) The first term G1 =2, the common ratio r=2 and n=8, Then from

Gn = G1rn-1

G8 =(2)(2)8-1

G8 = (2). 27

G8 = 256

(b) The first term G1 =12, the common ratio r=1/2 and n=8, Then from

Gn = G1rn-1

G8 =(12)(1/2)8-1

G8 = (12). (1/2)7

G8 = 12/128 = 3/32

Example2: Find the numbers of terms in the following 1+2+4+8+16+…+512.

Solution

The first term G1 =1, the common ratio r=2 and Gn= 512, Then from

Gn = G1rn-1

512 =(1)(2)n-1

512 = 2n-1

512 = 2n.2-1

512 x 2= 2n

1024= 2n

210 = 2n

n = 10

The sum of the first n terms of a geometrical progression.

Let the sum of first n terms of a G.P be denote by Sn

Sn = G1+ G2+ G3+ G4+…+ Gn

Since the common ratio is r

From

G2 = G1r

G3 = G1r2

G4 = G1r3

…Gn = G1rn-1

Sn = G1+ G1r+ G1r2+ G1r3+…+ Gnrn-1

Multiplying by common ratio r both sides we have

rSn