SEQUENCE

A sequence is an ordered list of numbers whose subsequent values are formed based on a definite rule. The numbers in the sequence are called terms and these terms are normally separated from each other by commas.

Examples:

2, 4, 6, 8, 10,……

Rule: Addition of 2 for subsequent terms.

70, 66, 62, 58, 54,……

Rule: Subtraction of 4 for subsequent terms.

3, -6, 12, -24,……

Rule: Multiply each term by –2.

Sequences are either finite or infinite

A finite sequence is a sequence whose terms can be counted. i.e. it has an end. These types of sequences are usually terminated with a full stop. e.g. (i) 3,5,7,9,11,13. (ii) -7,-10,-13,-16,-19,-21.

If however, the terms in the sequence have no end, the sequence is said to be infinite. These types of sequences are usually ended with three dots, showing that it is continuous. e.g.

(i) 5,8,11,14,17,20…

(ii) -35,-33,-31,-29,-27,…

TYPES OF SEQUENCE

We two types of sequences. They are:

(i) Arithmetic progression

(ii) Geometric progression

ARITHMETIC PROGRESSION (AP){LINEAR SEQUENCE}

If in a sequence of terms T1, T2, T3, ...Tn-1, Tn the difference between any term and the one preceding it is constant, then the sequence is said to be in arithmetic progression (A.P) and the difference is known as the common difference, denoted by d.

d = Tn – Tn-1, where n = 1, 2, 3, 4, …

i.e d = T2 – T1 = T3 – T2 = T4 – T3 and so on.

Examples of A.P

(i) 1, 3, 5,7, 9, …

Tn – Tn-1Ãž 5 - 3 = 2

7 - 5 = 2

9 - 7 = 2

d = 2

The difference is common, hence it is an A.P.

(ii) 2, 4, 8, 16, 32, …

Tn – Tn-1Ãž 4 - 2 = 2

8 - 4 = 4

16 - 8 = 8

32 - 16 = 16

The difference is NOT common; therefore it is not an A.P.

(iii) 70, 66, 62, 58, 54, …

Tn – Tn-1Ãž 66 - 70 = -4

62 - 66 = -4

58 - 62 = -4

d = -4

The difference is common; hence it is an A.P.

(iv) –2, -5, -8, -11, …

Tn – Tn-1Ãž (-5) - (-

2) = -5 + 2 = -3

(-8) - (-5) = -8 + 5 = -3

(-11) - (-8) = -11 + 8 = -3.

The difference is common; hence it is an A.P.

EXERCISE

Which of the following are arithmetic progressing sequence

1. 4,6,8,10,…

2. 3,7,9,11,..

3. 1,6,11,16,21,26…

4. 100,96,92,88,84,…

5. 20,17,15,11,…

6. 45,42,39,36,…

THE nth TERM OF AN A.P

If the first term of an A.P is 3 and the common difference is 2. The terms of the sequence are formed as follows.

1st term = 3

2nd term = 3+2 = 3 + (1)2

3rd term = 3+2+2 = 3 + (2)2

4th term = 3+2+2+2 = 3 + (3)2

5th term = 3+2+2+2+2 = 3 + (4)2

nth term = 3+2+2+2+ … = 3 + (n - 1)2

Hence, the nth term (Tn) of an A.P whose first term is “a” and the common difference is “d”

Example :

Find the21st term of the A.P 3, 5, 7, 9, …

Solution

a = 3

d = 2

n = 21

Tn = a + (n – 1)d

T21 = 3 + (21 – 1)2

= 3 + 20 x 2

= 3 + 40

= 43.

Example :

Find the 27th term of the A.P

100, 96, 92, 88, …

Solution

a = 100

d = -4

n = 27

Tn = a + (n – 1)d

T27 = 100 + (27 – 1)(-4)

= 100 + 26 x –4

= 100 – 104

= -4.

HOW TO SOLVE FOR ‘n’, ‘a’ AND ‘d’

Example :

Find the value of n given that 77 is the nth term of an A.P 3½, 7, 10½, …

Solution:

a = 3½

d = 7 - 3½

d = 3½

Tn= 77

Tn = a + (n – 1)d

77 = 3½ + (n – 1)3½

77 = 3½ + 3½n – 3½

77 = 3½n

77 = 7/2n

7n = 77 x 2

n = 77 x 2

7

n = 11 x 2

n = 22.

Example 4:

The question in example 1can be reframed as follows to find a.

“What is the first term of an A.P whose 21st term is 43 and the common difference is 2 ?”

Solution:

T21 = 43

n = 21

d = 2

Tn = a + (n –1)d

43 = a + 20 x 2

43 = a + 40

a = 43 – 40

a = 3

Example :

The same example in 4 above can be reframed also as follows to find d.

“Find the common difference of an A.P given that 43 is the 21st term of the sequence and the first term is 3”.

Solution:

a = 3

T21 = 43

n = 21

Tn = a + (n – 1)d

43 = 3 + (21 – 1) d

43 = 3 + 20d

43 –3 = 20d

20d = 40

d = 40

20

d = 2.

EXERCISES

1. Find the 31st term of the sequence –7, -10, -13, -16, …

2. What is the 26th term of the A.P 5, 10, 15, 20, …?

3. Find the 20th term of the sequence 27, 24, 21, 18, …

4. Find the 18th term of the A. P 6, 12, 18, 24, …

5 .Find the 26th term of the A.P –16, -13, -10, -7, …

6. Find the 29th term of the A.P 43, 39, 35, 31, …

7. Find n given that 697 is the nth term of the A.P –3, 4, 11, 18, …

8. Find n given that –8 is the nth term of the A.P 82, 79, 76, 73, 70, …

9. Find n given that 63 is the nth term of the sequence –17, -13, -9, -5, …

10. Find the number of terms in an A.P given that 147 is the last term of the A.P whose first term is 6 and common difference is 3.

11. Find the first term of an A.P given that 124 is the 41st term of the A.P and 3 is the common difference.

13. Find the first term of an A.P given that –5 is the 26th term of the sequence and –3 is the common difference.

14. Given that 57 is the 27th term of an A.P whose common difference is 2, find the first term.

15. Given that 76 is the 21st term of an A.P whose first term is 16. Find the common difference.

16. Given that 67 is the 21st term of an A.P whose first term is 7, find the common difference.

17. Find the common difference of an A.P, given that the 27th term is 90 and the first term is –14.

FURTHER EXAMPLES

Example :

The first three terms of an A.P are

x, 3x + 1, and (7x - 4). Find the

(i) Value of x

(ii) 10th term

Solution:

(i) Recall that given an A.P T1, T2, T3

T2 – T1 = T3 – T2

Hence for, x, (3x + 1), (7x - 4)

(3x + 1) – x = (7x - 4) – (3x + 1)

3x+1 – x = 7x – 4 – 3x - 1

2x + 1 = 4x – 5

1+ 5 = 4x – 2x

2x = 6

x = 6/2

x = 3.

The sequence x, (3x + 1), (7x - 4) is

= 3, (3x3 + 1), (7x3 - 4)

= 3, 10, 17.

(ii) a = 3

n = 10

d = 7

Tn = a + (n – 1)d

T10 = 3 + (10 – 1)7

= 3 + 9x7

= 3 + 63

= 66.

Example :

The 6th term of an A.P is –10 and the 9th term is –28.

Find the (i) Common difference

(ii) First term

(iii) 26th term of the sequence.

Solution:

(i) T6 = -10 Tn = a + (n - 1)d

n = 6 -10 = a + (6 - 1)d

-10 = a + 5d ----------- (1)

T9 = -28 -28 = a + (9 - 1)d

n = 9 -28 = a + 8d ---------- (2)

Solve equation (1) and (2) simultaneously.

Eqn. (1): -10 = a + 5d

Eqn. (2): -28 = a + 8d

18 = -3d

d =18/-3

d = -6.

(ii) Put d = -6 in equation (1)

-10 = a + 5(-6)

-10 = a - 30

-10+30 = a

a = 20.

(iii) To find the 26th term of the sequence.

a = 20

d = -6

n = 26

Tn= a + (n - 1)d

T26 = 20 + (26 - 1)(-6)

= 20 + 25(-6)

= 20 - 150

= -130.

EXERCISE

(1) The 6th term of an AP is –10 and the 9th term is 18 less than the 6th term. Find the

(a)common difference

(b) first term

(c) 26th term of the sequence.

(2) The 7th term of an AP is 17 and the 13th term is 12 more than the 7th term. Find the

(i) common difference

(ii) first term

(iii) 21st term of the AP.

(3) The 6th term of an A.P is 26 and the 11th term is 46. Find the

(i) Common difference

(ii) First term

(iii) 25th term of the A.P

(4) The 5th term of an A.P is 11 and the 9th term is 19. Find the (i) common difference

(ii) First term

(iii) 21st term of the A.P

(5) The fourth term of an A.P is 37 and the 6th term is 12 more than the fourth term. Find the first and seventh terms.

(6) The first three terms of an A.P are (x+2) ,(2x-5) and (4x+1). Find the

(i) Value of x

(ii) 7th term.

(7) The first three terms of an A.P are x, (2x-5) and (x+6). Find the

(i) Value of x

(ii) 21st term.

(8) If the first three terms of an A.P are (4x+1), (2x-5) and (x+3). Find the

(i) Value of x

(ii) Sequence

(iii) 11th term of the sequence

(9) Given that 9, x, y, 24 are in A.P, find the values of x and y.

(10) If –5, a, b, 16 are in A.P, find the values of a and b.

Arithmetic Series

These are series formed from an arithmetic progression. e.g.

1 + 4 + 7 + 10 + …

In general, if Sn is the sum of n terms of an arithmetic series then

Sn = a + (a + d) + (a + 2d) + … + (l - d) +

l -- (1)

Where l is the nth term, a is the first term and d is the common difference. Rewriting the series above starting with the nth term, we have.

Sn = l +(l - d) + (l - 2d) +… + (a + d) + a ----- (2)

Adding equation (1) and (2) we have

2Sn = (a + l) + (a + l) + … + (a + l) + (a + l) in n places

2Sn = n(a + l)

Sn = n/2(a + l)

But l is the nth term i.e a + (n - 1) d

Sn= n/2{a +a + (n - 1)d}

Sn = n/2 {2a + (n - 1)d}

Example :

Find the sum of the first 20 terms of the series 3+5+7+9+ …

Solution:

a = 3

d = 2

n = 20

Sn = n/2 {2a +(n - 1)d}

S20 = 20/2 {2x3 +(20 - 1)2}

S20 = 10{6 + 19 x 2}

= 10{6 + 38}

= 10{44}

S20 = 440

Example :

Find the sum of the first 28 terms of the series –17 + (-14) + (-11) + (-8) + …

Solution:

a = -17

d = 3

n = 28

Sn = n/2{2a + (n - 1)d}

S28 = 28/2 {2 (-17) + (28 - 1) 3}

= 14 {-34 + 27 x 3}

= 14 { -34 + 81}

= 14 {47}

S28 = 658

A sequence is an ordered list of numbers whose subsequent values are formed based on a definite rule. The numbers in the sequence are called terms and these terms are normally separated from each other by commas.

Examples:

2, 4, 6, 8, 10,……

Rule: Addition of 2 for subsequent terms.

70, 66, 62, 58, 54,……

Rule: Subtraction of 4 for subsequent terms.

3, -6, 12, -24,……

Rule: Multiply each term by –2.

Sequences are either finite or infinite

A finite sequence is a sequence whose terms can be counted. i.e. it has an end. These types of sequences are usually terminated with a full stop. e.g. (i) 3,5,7,9,11,13. (ii) -7,-10,-13,-16,-19,-21.

If however, the terms in the sequence have no end, the sequence is said to be infinite. These types of sequences are usually ended with three dots, showing that it is continuous. e.g.

(i) 5,8,11,14,17,20…

(ii) -35,-33,-31,-29,-27,…

TYPES OF SEQUENCE

We two types of sequences. They are:

(i) Arithmetic progression

(ii) Geometric progression

ARITHMETIC PROGRESSION (AP){LINEAR SEQUENCE}

If in a sequence of terms T1, T2, T3, ...Tn-1, Tn the difference between any term and the one preceding it is constant, then the sequence is said to be in arithmetic progression (A.P) and the difference is known as the common difference, denoted by d.

d = Tn – Tn-1, where n = 1, 2, 3, 4, …

i.e d = T2 – T1 = T3 – T2 = T4 – T3 and so on.

Examples of A.P

(i) 1, 3, 5,7, 9, …

Tn – Tn-1Ãž 5 - 3 = 2

7 - 5 = 2

9 - 7 = 2

d = 2

The difference is common, hence it is an A.P.

(ii) 2, 4, 8, 16, 32, …

Tn – Tn-1Ãž 4 - 2 = 2

8 - 4 = 4

16 - 8 = 8

32 - 16 = 16

The difference is NOT common; therefore it is not an A.P.

(iii) 70, 66, 62, 58, 54, …

Tn – Tn-1Ãž 66 - 70 = -4

62 - 66 = -4

58 - 62 = -4

d = -4

The difference is common; hence it is an A.P.

(iv) –2, -5, -8, -11, …

Tn – Tn-1Ãž (-5) - (-

2) = -5 + 2 = -3

(-8) - (-5) = -8 + 5 = -3

(-11) - (-8) = -11 + 8 = -3.

The difference is common; hence it is an A.P.

EXERCISE

Which of the following are arithmetic progressing sequence

1. 4,6,8,10,…

2. 3,7,9,11,..

3. 1,6,11,16,21,26…

4. 100,96,92,88,84,…

5. 20,17,15,11,…

6. 45,42,39,36,…

THE nth TERM OF AN A.P

If the first term of an A.P is 3 and the common difference is 2. The terms of the sequence are formed as follows.

1st term = 3

2nd term = 3+2 = 3 + (1)2

3rd term = 3+2+2 = 3 + (2)2

4th term = 3+2+2+2 = 3 + (3)2

5th term = 3+2+2+2+2 = 3 + (4)2

nth term = 3+2+2+2+ … = 3 + (n - 1)2

Hence, the nth term (Tn) of an A.P whose first term is “a” and the common difference is “d”

Example :

Find the21st term of the A.P 3, 5, 7, 9, …

Solution

a = 3

d = 2

n = 21

Tn = a + (n – 1)d

T21 = 3 + (21 – 1)2

= 3 + 20 x 2

= 3 + 40

= 43.

Example :

Find the 27th term of the A.P

100, 96, 92, 88, …

Solution

a = 100

d = -4

n = 27

Tn = a + (n – 1)d

T27 = 100 + (27 – 1)(-4)

= 100 + 26 x –4

= 100 – 104

= -4.

HOW TO SOLVE FOR ‘n’, ‘a’ AND ‘d’

Example :

Find the value of n given that 77 is the nth term of an A.P 3½, 7, 10½, …

Solution:

a = 3½

d = 7 - 3½

d = 3½

Tn= 77

Tn = a + (n – 1)d

77 = 3½ + (n – 1)3½

77 = 3½ + 3½n – 3½

77 = 3½n

77 = 7/2n

7n = 77 x 2

n = 77 x 2

7

n = 11 x 2

n = 22.

Example 4:

The question in example 1can be reframed as follows to find a.

“What is the first term of an A.P whose 21st term is 43 and the common difference is 2 ?”

Solution:

T21 = 43

n = 21

d = 2

Tn = a + (n –1)d

43 = a + 20 x 2

43 = a + 40

a = 43 – 40

a = 3

Example :

The same example in 4 above can be reframed also as follows to find d.

“Find the common difference of an A.P given that 43 is the 21st term of the sequence and the first term is 3”.

Solution:

a = 3

T21 = 43

n = 21

Tn = a + (n – 1)d

43 = 3 + (21 – 1) d

43 = 3 + 20d

43 –3 = 20d

20d = 40

d = 40

20

d = 2.

EXERCISES

1. Find the 31st term of the sequence –7, -10, -13, -16, …

2. What is the 26th term of the A.P 5, 10, 15, 20, …?

3. Find the 20th term of the sequence 27, 24, 21, 18, …

4. Find the 18th term of the A. P 6, 12, 18, 24, …

5 .Find the 26th term of the A.P –16, -13, -10, -7, …

6. Find the 29th term of the A.P 43, 39, 35, 31, …

7. Find n given that 697 is the nth term of the A.P –3, 4, 11, 18, …

8. Find n given that –8 is the nth term of the A.P 82, 79, 76, 73, 70, …

9. Find n given that 63 is the nth term of the sequence –17, -13, -9, -5, …

10. Find the number of terms in an A.P given that 147 is the last term of the A.P whose first term is 6 and common difference is 3.

11. Find the first term of an A.P given that 124 is the 41st term of the A.P and 3 is the common difference.

13. Find the first term of an A.P given that –5 is the 26th term of the sequence and –3 is the common difference.

14. Given that 57 is the 27th term of an A.P whose common difference is 2, find the first term.

15. Given that 76 is the 21st term of an A.P whose first term is 16. Find the common difference.

16. Given that 67 is the 21st term of an A.P whose first term is 7, find the common difference.

17. Find the common difference of an A.P, given that the 27th term is 90 and the first term is –14.

FURTHER EXAMPLES

Example :

The first three terms of an A.P are

x, 3x + 1, and (7x - 4). Find the

(i) Value of x

(ii) 10th term

Solution:

(i) Recall that given an A.P T1, T2, T3

T2 – T1 = T3 – T2

Hence for, x, (3x + 1), (7x - 4)

(3x + 1) – x = (7x - 4) – (3x + 1)

3x+1 – x = 7x – 4 – 3x - 1

2x + 1 = 4x – 5

1+ 5 = 4x – 2x

2x = 6

x = 6/2

x = 3.

The sequence x, (3x + 1), (7x - 4) is

= 3, (3x3 + 1), (7x3 - 4)

= 3, 10, 17.

(ii) a = 3

n = 10

d = 7

Tn = a + (n – 1)d

T10 = 3 + (10 – 1)7

= 3 + 9x7

= 3 + 63

= 66.

Example :

The 6th term of an A.P is –10 and the 9th term is –28.

Find the (i) Common difference

(ii) First term

(iii) 26th term of the sequence.

Solution:

(i) T6 = -10 Tn = a + (n - 1)d

n = 6 -10 = a + (6 - 1)d

-10 = a + 5d ----------- (1)

T9 = -28 -28 = a + (9 - 1)d

n = 9 -28 = a + 8d ---------- (2)

Solve equation (1) and (2) simultaneously.

Eqn. (1): -10 = a + 5d

Eqn. (2): -28 = a + 8d

18 = -3d

d =18/-3

d = -6.

(ii) Put d = -6 in equation (1)

-10 = a + 5(-6)

-10 = a - 30

-10+30 = a

a = 20.

(iii) To find the 26th term of the sequence.

a = 20

d = -6

n = 26

Tn= a + (n - 1)d

T26 = 20 + (26 - 1)(-6)

= 20 + 25(-6)

= 20 - 150

= -130.

EXERCISE

(1) The 6th term of an AP is –10 and the 9th term is 18 less than the 6th term. Find the

(a)common difference

(b) first term

(c) 26th term of the sequence.

(2) The 7th term of an AP is 17 and the 13th term is 12 more than the 7th term. Find the

(i) common difference

(ii) first term

(iii) 21st term of the AP.

(3) The 6th term of an A.P is 26 and the 11th term is 46. Find the

(i) Common difference

(ii) First term

(iii) 25th term of the A.P

(4) The 5th term of an A.P is 11 and the 9th term is 19. Find the (i) common difference

(ii) First term

(iii) 21st term of the A.P

(5) The fourth term of an A.P is 37 and the 6th term is 12 more than the fourth term. Find the first and seventh terms.

(6) The first three terms of an A.P are (x+2) ,(2x-5) and (4x+1). Find the

(i) Value of x

(ii) 7th term.

(7) The first three terms of an A.P are x, (2x-5) and (x+6). Find the

(i) Value of x

(ii) 21st term.

(8) If the first three terms of an A.P are (4x+1), (2x-5) and (x+3). Find the

(i) Value of x

(ii) Sequence

(iii) 11th term of the sequence

(9) Given that 9, x, y, 24 are in A.P, find the values of x and y.

(10) If –5, a, b, 16 are in A.P, find the values of a and b.

Arithmetic Series

These are series formed from an arithmetic progression. e.g.

1 + 4 + 7 + 10 + …

In general, if Sn is the sum of n terms of an arithmetic series then

Sn = a + (a + d) + (a + 2d) + … + (l - d) +

l -- (1)

Where l is the nth term, a is the first term and d is the common difference. Rewriting the series above starting with the nth term, we have.

Sn = l +(l - d) + (l - 2d) +… + (a + d) + a ----- (2)

Adding equation (1) and (2) we have

2Sn = (a + l) + (a + l) + … + (a + l) + (a + l) in n places

2Sn = n(a + l)

Sn = n/2(a + l)

But l is the nth term i.e a + (n - 1) d

Sn= n/2{a +a + (n - 1)d}

Sn = n/2 {2a + (n - 1)d}

Example :

Find the sum of the first 20 terms of the series 3+5+7+9+ …

Solution:

a = 3

d = 2

n = 20

Sn = n/2 {2a +(n - 1)d}

S20 = 20/2 {2x3 +(20 - 1)2}

S20 = 10{6 + 19 x 2}

= 10{6 + 38}

= 10{44}

S20 = 440

Example :

Find the sum of the first 28 terms of the series –17 + (-14) + (-11) + (-8) + …

Solution:

a = -17

d = 3

n = 28

Sn = n/2{2a + (n - 1)d}

S28 = 28/2 {2 (-17) + (28 - 1) 3}

= 14 {-34 + 27 x 3}

= 14 { -34 + 81}

= 14 {47}

S28 = 658