## 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 **

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 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

3.

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

d=

From the 1^{st} equation

A_{1} += 2

A_{1} = 2+

A_{1} =

4.

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

5.

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

6.

(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.

7.

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

8.

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

4+

X =

A_{3} = A_{1}+2d

4+ 2x

A_{3} =

Hence x= and y=

9.

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 termA_{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)**

Definition:

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 = G1r2
G4 = G1r3
.}^{.}_{.Gn = G1rn-1
Sn = G1+ G1r+ G1r2+ G1r3+…+ Gnr}^{n-1}_{
Multiplying by common ratio r both sides we have}

rSn = rG1+ _{+ G1r3+ G1r4+…+ Gnr}^{n}_{}

Substract Sn from rSn

_{rSn – Sn = }_{ – G1
}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=

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_{3}= 3 + A_{1} + 2d

S_{10} =?

From,A_{1}+ A_{4}= 18

A_{1}+ A_{1}+ 3d =18

2A_{1}+ 3d = 18………..(i) and

A_{1 }+ 4d =3 + A_{1} + 2d……..(ii)

Take equation (ii)

A_{1} + 4d = 3 + A_{1 }+ 2d

2d= 3

_{
d = 3/2}

Substitute the value of d into equation …….. (i)

2A_{1}+ 3d = 18

2A_{1}+ 3(3/2) = 18

2A_{1} + 9/2 = 18

2A_{1}= 18 – 9/2 = 27/2

2A_{1}= (27/2) Divide by 2 both sides

A_{1}= 27/4

So, The sum of the first ten terms S_{n} = n/2 (A_{1} + A_{n})

S_{n} = n/2 (A_{1}+ A_{1} + (n – 1)d)

S_{n} = n/2 (2A_{1}+ (n – 1)d)

S_{10} = 10/2 ((27/4)2+ (10 – 1) x 3/2)

S_{10} = 135

Therefore the value of ten terms will be,S_{10} = 135

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} =

S_{n} =

10430 =

5215 =2^{n}-1

5216 =2^{n} then

n=

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

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=

p=

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.

**5.7. APPLICATION OF A.P AND G.P.**

Simple interest is an application of arithmetic progression which is given by;

I =

**Where **

I=simple interest

P=principal

R= rate of interest

T= period of interest

Compound interest is an application of geometric progression it is given by;

A_{n}=p+1 or A_{n} =p)^{n}

**Where **

A_{n} =an amount at the end of the New Year

R= rate of interest

n= number of years

T=period of interest

p= principal

**Example**

1. Find the simple interest on Tshs 10,000/= deposited in a bank at the rate of 10% annually for 4 years

**Solution**

I =

I =

I=1000×4

I=4000

The interest for 4 years will be 4000/=