Form 3 Mathematics – 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[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 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 = 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

A_n = A_[n-1] + d = A₁+ (n-1)d

.^.. The general formula for obtaining the nth term of the series is
A_n = A₁+ (n-1)d

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

Question

1. A₇ = A_{2 and} A₄ =16. Find A₁ and d.

Solution

A₇= A₁+6d
=A₁+6d=  (  A₁+d)

2A₁+12d=5A₁+5d

        3A₁=7d
A₁= d

A ₄ =16
A₄=A₁+3d = 16
 d +3d= 16

        d = 16

         d= 3

But A₁ +3d= 16

A₁ +9=16
A₁= 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 = 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]

                 r^th=A₁+[y-x][n-1]

2.

A₅=17

A₃=11
A₁₃=?

A₅ =A₁+4d = 17
A₃=A₁+2d =17
A₁+2d=11

Solve the simultaneous equation by using equilibrium method.

A₁+4d=17
A₁+2d=11
  =  

      d = 3

A₁+4[3]=17
A₁+12=17-12
A₁=5

A₁₃=A₁+12d
A₁₃=5+12[3]
A₁₃=5+36
A₁₃ =41

3. 

A₂=2
A₁₆=-4
A₁=?

A₂=A₁+d
A₁+d=2

A₁₆=A₁+15d
A₁ +15d= -4

Solve the two simultaneously  equations by using the elimination method

  A₁  +   d=2

  A₁ +15d=-4

    d=    

From the 1^st equation

A₁ + = 2

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 1^st equation

A₁ +5[2] = 14
A₁ +10= 14-10
A₁=4

A₁₀= A₁+9d
A₁₀=4+9[2]
A ₁₀=4+18
A₁₀=22

A_n =A₁+[n-1]d
A_n =4[n-1]2
A_n =4+2n-2
A_n= 2n+4-2
A_n=2n+2

5.

A₂= 3 x A_b
D= -4
A₁=?
A_n=?

A_z  =  3 x A₆
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

A_n = A₁+[n-1]d
A_n=  28+[n-1]-4
A_n=32-4n
A_n =32-4n

6. 

(a) A₃= 0
d= -2

     From the formula

     A₁+2d= 0
A₁+2[-2]= 0
A₁-4=0
A₁= 4

     The first term is 4.

b)      The general term

A_n =A₁+ [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.

A₅₄ =?
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

8.

4, x, y, 20
A ₄= 20
A₁ +3d=20 , but A₁=4

4+3d= 20
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=?

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

A₁ +39[3]
A₁+117
4+117
A₄₀ =121

10.

A₁ =4
A₂=9
A₃=14
d= 9-4= 14-9
d= 5

A_n =A₁+[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
A₅= 40
A₇= 20

A₁+ 4d= 40         ………… (1)
A₁+ 6d= 20         ………….(2)

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

-2d=20
d= -10

   b)  the tenth term

A₁₀= A₁+9d,
But A₁ +4d=40
A₁=80

∴A₁₀=A₁+9d
=80-90
A ₁₀=-10

12.

A₂= A₁+d
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 =

 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 S_n, then

S_n = A₁ + (A₁+ d) + (A₁ + 2d)+ …+ (A₁ – d) + A_n
+  S_n = A_n + (A_n- d) + (A_n – 2d)+ …+ (A₁ + d) + A₁
                                                                                                                     

2S_n = (A₁+ A_n) + (A₁+ A_n) + (A₁+ A_n)+ …+ (A₁+ A_n)+ (A₁+ A_n)

There are n terms of (A₁+ A_n) then
2S_n = n(A₁+ A_n)
S_n = n(A₁+ A_n)
2
.^.. The sum of the first n terms of an A.P with first term A₁ and the last termA_n is given by
S_n = (A₁+ A_n)

But A_n =A₁+ (n-1) d
Thus, from
S_n = (A₁+ A_n)
S_n = [A₁+ A₁+ (n-1) d]
S_n = [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
S_n = [2A₁+ (n-1) d]

Where

n =number of terms
A₁= 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

S₅ = [2 + 14]
S₅ = 40

  (ii)S_n = [2A₁ + [n-1] d]
S₅ = [2 x 2 + [5-1] (3)]

       S₅= 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.
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 n^th term is given by
G_n = G₁r^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   G₁ =2, the common ratio r=2 and n=8, Then from
G_n = G₁r^n-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
G_n = G₁r^n-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 G_n= 512, Then from
G_n = G₁r^n-1
512 =(1)(2)^n-1
512 = 2^n-1
       512 = 2ⁿ.2^-1
512 x 2= 2ⁿ
1024= 2ⁿ
2¹⁰ = 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 = G₁+ G₂+ G₃+ G₄+…+ 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) = (rⁿ– 1)
S_n =     where r ≥ 1     for r≠1

G₁= 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 = G₁ – G₁rⁿ

S_n(1-r) = G₁(1- rⁿ)

S_n=  for r<1

When r= 1, the sum is simply given by
S_n= G₁+G₁+G₁+G₁+G₁+G₁+…+G₁

   s_n=nG₁ for r=1

Example

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

S₅ =
S₅ =    

 S₅ = 62

S_n =    

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

    Solution

     When

   n= 1;
S₁ = 3¹-1 = 3  – 1
= 2

n= 2;
S₂ = 3²-1 = 9 – 1
= 8

   n= 3;
S₃ =  3³-1  = 27  – 1
= 26

   n= 4;
S₄ =  3⁴-1  = 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

S₅ = 35
A₅ =395
d= ?
A₁ =?

S₅ = [2A₁ + [5-1] d]
S₅ = [2A₁+4d]

S₅= 5A₁+10d

35= 5A₁+10d … (1)
395=5 A₁+190d… (2)

Solve the simultaneous equations

Then the value of d = 2

From the 1^st equation

5A₁+10d=35
5A₁+10[2]=35
5A₁+ 20 =35
A₁ = 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

A₁ =4
A_n =256
S_n =1820
n=?
d=?
S_n = [A₁ + A_n]

1820 =  [4 + 256]

1820 = n [130]

     n= 

     n= 14

Therefore the n term = 14

An = A₁+ [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

A₄= 2x+10
A₅= 40x-4
A₆=8x+40

Solution

A₄= A₁+3d = (2x + 10)…i
A₅ = A₁+4d = (40x – 4)…ii

Solve the equations by using elimination method

A₁+3d= 2x+10
A₁+4d= 40x-4

– d = 2x + 10 – 40x + 4
d= 38x – 14

From 1^st equation

A₁+3[38x-14] =2x+10
A₁+114x – 42=2x +10
A₁ = 2x+10 – 114x + 42
A₁= -112x +52

Therefore the first terms = -112x+52

S₁₀= [ 2(-112x+52)  +[10-1] 38x-14]

S₁₀=5[-224x+104] +9 [38x-14]

S₁₀=5[-224x+104+342x-126]

S₁₀=5[118x-22]

S₁₀=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

A₁+A₄=18
A₅  = ( A₁  + 4d )=3( A₁ + 2d)
S₁₀ =?
A₁+A₁+3d =18

2A₁+3d=18………..i     and ( A₁  + 4d )=3( A₁ + 2d)

A₁  + 4d  = 3 A₁  + 6d
2 A₁   = – 2d
-A_{1 =}  d

Substitute the value of d into equation … (i)

2A₁+3d=18
2A₁+3(-A₁)=18
A₁ = -18

But  -A_{1 =} d            this give us d =  18

The sum of the first ten terms      S_n = [2A₁ + [n-1] d

                                                    S₁₀ = [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

G₁+G₂+G₃+G₄…………. 2+4+8+16

S_n =    

S_n =    

 10430 =    

   5215 =2ⁿ-1

5216 =2ⁿ  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

G₃= G₁ r² = 18 ………… (1)
G₆= G₁ r⁵ = 486……….. (2)

Take equation (2) divide by equation (1)

       =

      = 27

             r= 3

        But   = 18

                    

          G₁  =  2

   S_n =   then

     S₁₀ =    

         = 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=p²-2p+1=3p²-5p-6p+10

p²-2p+1=3p²-11p+10

p²-2p+1=3p²-11p+10

0=3p ²-p²-11p+2p+10-1

2p²-9p+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
M² = 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/=