Sequence And Series In Mathematics Pdf

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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

Is a set of number with a simple pattern.

Example

• 2, 4, 6, 8, 10 ……

• 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 3 rd term =6

iii)The 5 th term =10

iv)The n th term [ the general formula]

2=2×1

4=2×2

6=2×3

8=2×4

10=2×5

12=2×6

N th =2xn

Therefore n th term =2n

Find the 100 th term, general formula =2n

100 th term means n=100

100 th term =2×100

100 th term =200

3. Find the n th term

Solution

4. Given the general term 3[2 n ]

a) Find the first 5terms

b) Find the sum of the first 3 terms

Solution

3[2 n ] when;

n=1, 3 [2 1 ] =6

n=2, 3 [2 2 ] =12

n=3, 3 [2 3 ] =24

n=4, 3 [2 4 ] =48

n=5, 3 [2 5 ] =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 n th term of a sequence is given by 2n+1 write down the 10 th term.
4. The n th term of a certain sequence is 2 n-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…n th

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

n th =2+n

2. 3, 6, 9, 12……..n th

The difference between the consecutive terms is 3 thus

n th =3n

3. 10 th = 2[10+1]

10 th =20+1

10 th =21

4. When

n=1, 2 1-1

n=2, 2 2-1

n=3, 2 3-1

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 n th 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 n th 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 n th term of an arithmetic progression
If n is the number of terms of an arithmetic progression, then the n th term is denoted by A n
Therefore A n+1 = A n + d
e.g. First term = A 1
Second term = A 1 + d = A 2
Third term = A 2 + d = A 3

Consider a series 3+6+9+12+…..
A 1 = 3, d= 3
A 2 = A 1 + d
A 3 = A 2 + d = A 1 +d+d = A 1 + 2d
A 4 = A 3 + d = A 1 + 2d+ d = A 1 + 3d
A 5 = A 4 + d = A 1 + 3d+ d = A 1 + 4d
A 6 = A 5 + d = A 1 + 4d+ d = A 1 + 5d

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

. . . The general formula for obtaining the nth term of the series is
A n = A 1 + (n-1)d

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

Question

1. A 7 = A 2 and A 4 =16. Find A 1 and d.

Solution

A 7 = A 1 +6d
=A 1 +6d= ( A 1 +d)

2A 1 +12d=5A 1 +5d

3A 1 =7d
A 1 = d

A 4 =16
A 4 =A 1 +3d = 16
d +3d= 16

d = 16

d= 3

But A 1 +3d= 16

A 1 +9=16
A 1 = 7

Exercise 2

1. The p th term of an A.P is x and the q th term of this is y, find the r th term of the same A.P
2. The fifth term of an A.P is 17 and the third term is 11. Find the 13 th term of this A.P.
3. The second term of an A.P is 2 and the 16 th term is -4 find the first term.
4. The sixth term of an A.P is 14 and the 9 th of the same A.P is 20 find 10 th term.
5. The second term of an A.P is 3 times the 6 th term. If the common difference is -4 find the 1 st term and the n th 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 54 th 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 40 th term of an A.P 4, 7, 10……..
10.What is the n th term of an A.P 4, 9, 14

11.The 5 th 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 n th term
12. The 2 nd term of an A.P is 7 and the 7 th term is 10 find the first term and the common difference

Exercise 2 Solution

1. d = y-x
r th = A 1 +[n-1]d

= A 1 +nd-d

= A 1 +n[y-x]-[y-x]

= A 1 +[ny-nx-[y-x]

=A 1 +[ny-y]-[nx-x]

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

r th =A 1 +[y-x][n-1]

2 .

A 5 =17

A 3 =11
A 13 =?

A 5 =A 1 +4d = 17
A 3 =A 1 +2d =17
A 1 +2d=11

Solve the simultaneous equation by using equilibrium method.

A 1 +4d=17
A 1 +2d=11
=

d = 3

A 1 +4[3]=17
A 1 +12=17-12
A 1 =5

A 13 =A 1 +12d
A 13 =5+12[3]
A 13 =5+36
A 13 =41

A 2 =2
A 16 =-4
A 1 =?

A 2 =A 1 +d
A 1 +d=2

A 16 =A 1 +15d
A 1 +15d= -4

Solve the two simultaneously equations by using the elimination method

A 1 + d=2

A 1 +15d=-4

From the 1 st equation

A 1 + = 2

A 1 = 2+

A 1 =

A 6 = 14

A 9 =20

A 6 = A 1 +5d = 14

A 9 = A 1 +8d=20

Solve the simultaneous equation by using the elimination method

A 1 +5d= 14

A 1 +8d=20 ( difference between two equations)

Then d = 2

From 1 st equation

A 1 +5[2] = 14
A 1 +10= 14-10
A 1 =4

A 10 = A 1 +9d
A 10 =4+9[2]
A 10 =4+18
A 10 =22

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

A 2 = 3 x A b
D= -4
A 1 =?
A n =?

A z = 3 x A 6
A 1 +d = 3 x A1+5d
A 1 +d =3A1+15d
2A 1 +14d=0
d= -4

2A 1 +14[-4]=0
2A 1 +-56= 0
2A 1 = 56
A 1 = 28

A n = A 1 +[n-1]d
A n = 28+[n-1]-4
A n =32-4n
A n =32-4n

(a) A 3 = 0
d= -2

From the formula

A 1 +2d= 0
A 1 +2[-2]= 0
A 1 -4=0
A 1 = 4

The first term is 4.

b) The general term

A n =A 1 + [n-1] d
A n =4+[n-1]-2
A n = 4+ -2n+2
A n = 6-2n

The general term is 6-2n.

A 54 =?
100, 97, 94 = A 1 , A 2 , A 3

A 54 = A 1 +53d
d= A 2 -A 1 = A 3 -A 2
d= 97-100= 94-97
d= -3

A 54 = 100+53[-3]
A 54 =100 + -159
A 54 = -59

4, x, y, 20
A 4 = 20
A 1 +3d=20 , but A 1 =4

4+3d= 20
3d=16
d = 16 /3

A 1 , A 2 , A 3 , A 4

A 2 =A 1 +d

X =

A 3 = A 1 +2d

4+ 2x

A 3 =

Hence x= and y=

A 40 =?
A 1 =4
A 2 = 7
A 3 = 10
A 1 +39d=?

d= 7-4= 10-7
d= 3

A 1 +39[3]
A 1 +117
4+117
A 40 =121

10.

A 1 =4
A 2 =9
A 3 =14
d= 9-4= 14-9
d= 5

A n =A 1 +[n-1]d
A n =4+[n-1]5
A n =4+5n-5
A n =5n-1

The nth term is 5n-1.

11.

a) the common difference
A 5 = 40
A 7 = 20

A 1 + 4d= 40 ………… (1)
A 1 + 6d= 20 ………….(2)

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

-2d=20
d= -10

b) the tenth term

A 10 = A 1 +9d,
But A 1 +4d=40
A 1 =80

∴ A 10 =A 1 +9d
=80-90
A 10 =-10

12.

A 2 = A 1 +d
A 7 = A 1 +6d
A 1 +6d = 10
A 1 +d=7

Solving the simultaneous equations by using the elimination method;

-5d= -3
d = 3 /5
A 1 + = 7
A 1 =

SUM OF THE FIRST n TERMS OF AN ARITHMETIC PROGRESSION
Consider a series with first term A 1 , common difference d. If the sum n terms is denoted by S n , then

S n = A 1 + (A 1 + d) + (A 1 + 2d)+ …+ (A 1 – d) + A n
+ S n = A n + (A n- d) + (A n – 2d)+ …+ (A 1 + d) + A 1

2S n = (A 1 + A n ) + (A 1 + A n ) + (A 1 + A n )+ …+ (A 1 + A n )+ (A 1 + A n )

There are n terms of (A 1 + A n ) then
2S n = n(A 1 + A n )
S n = n(A 1 + A n )
2
. . . The sum of the first n terms of an A.P with first term A 1 and the last termA n is given by
S n = (A 1 + A n )

But A n =A 1 + (n-1) d
Thus, from
S n = (A 1 + A n )
S n = [A 1 + A 1 + (n-1) d]
S n = [2A 1 + (n-1) d]

. . . therefore, the sum of the first n term of an A.P with the first A 1 and the common difference d in given by
S n = [2A 1 + (n-1) d]

Where

n =number of terms
A 1 = first term
A n =last term
d= common difference

Example

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

Solution

S 5 = [2 + 14]
S 5 = 40

(ii)S n = [2A 1 + [n-1] d]
S 5 = [2 x 2 + [5-1] (3)]

S 5 = 40

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 n th term of Geometric Progression
If n is the number of terms of G.P, the n th term is denoted by G n and common ratio by r. Then G n+1 = rG n for all natural numbers.
G 1 = G 1
G 2 = G 1 r
G 3 = G 2 r = G 1 r.r = G 1 r 2
G 4 = G 3 r = G 1 r 2 .r = G 1 r 3
G 5 = G 4 r = G 1 r 3 .r = G 1 r 4
G 6 = G 5 r = G 1 r 4 .r = G 1 r 5
G 7 = G 6 r = G 1 r 5 .r = G 1 r 6
G 8 = G 7 r = G 1 r 6 .r = G 1 r 7
The n th term is given by
G n = G 1 r n-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 G 1 =2, the common ratio r=2 and n=8, Then from
G n = G 1 r n-1
G 8 =(2)(2) 8-1
G 8 = (2). 2 7
G 8 = 256
(b) The first term G 1 =12, the common ratio r=1/2 and n=8, Then from
G n = G 1 r n-1
G 8 =(12)(1/2) 8-1
G 8 = (12). (1/2) 7
G 8 = 12/128 = 3/32

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

Solution
The first term G 1 =1, the common ratio r=2 and G n = 512, Then from
G n = G 1 r n-1
512 =(1)(2) n-1
512 = 2 n-1
512 = 2 n .2 -1
512 x 2= 2 n
1024= 2 n
2 10 = 2 n
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 S n
S n = G 1 + G 2 + G 3 + G 4 +…+ G n
Since the common ratio is r
From
G2 = G1r
G3 = G1r 2
G4 = G1r 3
…G n = G 1 rn-1
S n = G 1 + G 1 r+ G 1 r2+ G 1 r3+…+ G n rn-1
Multiplying by common ratio r both sides we have

rS n = rG 1 + + G 1 r3+ G 1 r4+…+ G n rn

Substract S n from rS n

rS n – S n = – G 1
S n (r- 1) = (r n – 1)
S n = where r ≥ 1 for r≠1

G 1 = first term of G.P
r= common ratio
Sn = sum of the first n th term
n= number of terms

for r< 1 the formular is given by

s n – rs n = G 1 – G 1 r n

S n (1-r) = G 1 (1- r n )

S n = for r<1

When r= 1, the sum is simply given by
S n = G 1 +G 1 +G 1 +G 1 +G 1 +G 1 +…+G 1

s n =nG 1 for r=1

Example

1. Sum of the first 5 th term of G.P where series is 2+4+8+16+32

S 5 =
S 5 =

S 5 = 62

S n =

2. The sum of the first n terms of a certain series is given by Sn= 3 n -1 show that this series is a G.P

Solution

When

n= 1;
S 1 = 3 1 -1 = 3 – 1
= 2

n= 2;
S 2 = 3 2 -1 = 9 – 1
= 8

n= 3;
S 3 = 3 3 -1 = 27 – 1
= 26

n= 4;
S 4 = 3 4 -1 = 81 – 1
= 80

2 + 6 + 18 + 54…
r=

r= 3

Exercise

1. An arithmetic progression has 41 terms. The sum of the first five terms of this A.P is 35 and the sum of the last five terms of the same A.P is 395 find the common difference and the first term.

Solution

S 5 = 35
A 5 =395
d= ?
A 1 =?

S 5 = [2A 1 + [5-1] d]
S 5 = [2A 1 +4d]

S 5 = 5A 1 +10d

35= 5A 1 +10d … (1)
395=5 A 1 +190d… (2)

Solve the simultaneous equations

Then the value of d = 2

From the 1 st equation

5A 1 +10d=35
5A 1 +10[2]=35
5A 1 + 20 =35
A 1 = 3

Therefore the first term is = 3

2. An arithmetic progression has the first term of 4 and n th term of 256 given that the sum of the n th term is 1280. Find the value of the n th term and common difference

Solution

A 1 =4
A n =256
S n =1820
n=?
d=?
S n = [A 1 + A n ]

1820 = [4 + 256]

1820 = n [130]

n= 14

Therefore the n term = 14

An = A 1 + [n-1]d
An = 4+[14-1]d
256= 4+13d
256-4=13d
252 = 13d
d=

∴ Common difference =

3. The 4 th , 5 th and6 th terms of an A.P are (2x +10), (40x-4) and (8x+40) respectively. Find the first term and the sum of the first 10

A 4 = 2x+10
A 5 = 40x-4
A 6 =8x+40

Solution

A 4 = A 1 +3d = (2x + 10)…i
A 5 = A 1 +4d = (40x – 4)…ii

Solve the equations by using elimination method

A 1 +3d= 2x+10
A 1 +4d= 40x-4

– d = 2x + 10 – 40x + 4
d= 38x – 14

From 1 st equation

A 1 +3[38x-14] =2x+10
A 1 +114x – 42=2x +10
A 1 = 2x+10 – 114x + 42
A 1 = -112x +52

Therefore the first terms = -112x+52

S 10 = [ 2(-112x+52) +[10-1] 38x-14]

S 10 =5[-224x+104] +9 [38x-14]

S 10 =5[-224x+104+342x-126]

S 10 =5[118x-22]

S 10 =590x-110

4. The sum of the first n terms of an A.P is given by s n =n[n+3] for all integral values of n. write the first four terms of the series

Solution

When;

n = 1 then s n =n[n+3] =1[1+3]

= 4

n = 2 then =2[2+3]

= 10

n = 3 then =3[3+3]

= 18

n = 4 then =4[4+3]

= 28

The first four terms of the series is 4+6+8+10

5. The sum of the first and fourth terms of an A.P is 18 and the fifth terms is 3 more than the third term. Find the sum of the first 10 terms of this A.P

Solution

A 1 +A 4 =18
A 5 = ( A 1 + 4d )=3( A 1 + 2d)
S 10 =?
A 1 +A 1 +3d =18

2A 1 +3d=18………..i and ( A 1 + 4d )=3( A 1 + 2d)

A 1 + 4d = 3 A 1 + 6d
2 A 1 = – 2d
-A 1 = d

Substitute the value of d into equation … (i)

2A 1 +3d=18
2A 1 +3(-A 1 )=18
A 1 = -18

But -A 1 = d this give us d = 18

The sum of the first ten terms S n = [2A 1 + [n-1] d

S 10 = [2 x -18 + [10-1] (18)]

= 630

6. How many terms of the G.P 2+4+8+16…… must be taken to give the sum greater than 10,430?

Solution

G 1 +G 2 +G 3 +G 4 …………. 2+4+8+16

S n =

10430 =

5215 =2 n -1

5216 =2 n then

Then more than term should be taken to provide the sum greater than 10430

7. In a certain geometrical progression, the third term is 18 and the six term is 486, find the first term and the sum of the first 10 terms of this G.P

Solution

G 3 = G 1 r 2 = 18 ………… (1)
G 6 = G 1 r 5 = 486……….. (2)

Take equation (2) divide by equation (1)

= 27

r= 3

But = 18

G 1 = 2

S n = then

S 10 =

= 2046

The first term is 2

The sum of the first 10 terms is 2046

8. Given that p-2, p-1 and 3p-5 are three consecutive terms of geometric progression find the possible value of p

Solution

R= =

r=[p-1][p-1]=[p-2][3p-5]

r=p 2 -2p+1=3p 2 -5p-6p+10

p 2 -2p+1=3p 2 -11p+10

0=3p 2 -p 2 -11p+2p+10-1

2p 2 -9p+9=0

From the general quadratic formula

P=3 or p=

Geometric mean

If a, m and b are consecutive terms of a geometric progression then the common ratio
r= M/a = b/M
M 2 = ab
M =

Example: Find the geometric mean of 4 and 16.
from G. M =

G.M=

G.M =

G.M = 12.