Arithmetic Sequence and Series

 

Consider the following sequences.

 

a. 1, 6, 11, 16, ... … …

b. 6, 2, -2, -6, … … … …

c. 3, 4.5, 6, 7.5, … … …

 

In the first sequence, each term is increased by 5 than the preceding term. In the second sequence, each term is decreased by 4 than the preceding term. Each term in the third sequence is increased by 1.5 than the preceding term. In each of the above sequences, the difference between a term and its preceding term is equal or constant. Such a sequence is said to be an Arithmetic Sequence or Arithmetic Progression.

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Definition of Arithmetic Sequence and Series

 

A sequence is said to be an arithmetic sequence or arithmetic progression if the difference between a term and its preceding term is equal or constant throughout the whole sequence. It is denoted by A.P.

 

The constant difference obtained by subtracting a term from its succeeding term is called the common difference. In a sequence 3, 8, 13, 18, … … …, the common difference = 8 – 3 = 13 – 8 = 18 – 13 = 5. Similarly, the common difference of 40, 36, 32, 28, … … … is -4 and that of 3, 4.5, 6, 7.5, … … … is 1.5. The common difference of an arithmetic sequence is denoted by d.

 

A series corresponding to any arithmetic sequence is known as the arithmetic series associated with the given arithmetic sequence. Hence, 3 + 8 + 13 + 18 + … … … is an arithmetic series associated with the arithmetic sequence 3, 8, 13, 18, … … … .

 

General term of an A.P.

 

If a be the first term and d be the common difference of an arithmetic progression, then the terms of the progression is a, a + d, a + 2d, a + 3d, … … …

 

If t₁, t₂, t₃, t₄, … … …, t_n be the first, second, third, fourth, … … …, n^th term of an A.P., then

 

t₁ = a = a + (1 – 1)d

t₂ = a + d = a + (2 – 1)d

t₃ = a + 2d = a + (3 – 1)d

t₄ = a + 3d = a + (4 – 1)d

… … …

t_n = a + (n – 1)d

 

Thus if the first term and the common difference of an A.P. are known, we can find any term by using the above formula.

 

 

Worked Out Examples

 

Example 1: What is the 7^th term of an A.P., when its first term and the common difference are 7 and 3 respectively.

 

Solution:

 

Here,

 

First term (a) = 7

Common difference (d) = 3

7^th term (t₇) = ?

 

We have,

 

t_n = a + (n – 1)d

∴    t₇ = 7 + (7 – 1)3

          = 7 + 6 × 3

          = 7 + 18

          = 25

 

Example 2: If 15^th term of an A.P. with first term 8 is 92, find the common difference.

 

Solution:

 

Here,

 

15^th term (t₁₅) = 92

First term (a) = 8

Common difference (d) = ?

 

By using the formula, we have

 

t_n = a + (n – 1)d

i.e. t₁₅ = 8 + (15 – 1)d

or, 92 = 8 + 14d

or, 14d = 92 – 8

or, 14d = 84

or, d = 84/14

or, d = 6

 

∴ The common difference = 6

 

 

Example 3: If three numbers 2k – 3, 3k, and 5k – 2 are in A.P.

a. Find the value of k.

b. Find the three numbers in A.P.

 

Solution:

 

a. As the three numbers 2k – 3, 3k, and 5k – 2 are in A.P.

3k – (2k – 3) = 5k – 2 – 3k

or, 3k – 2k + 3 = 2k – 2

or, k + 3 = 2k – 2

or, 3 + 2 = 2k – k

or, k = 5

∴ k = 5

 

b. The three numbers are 2k – 3, 3k, and 5k – 2

or, 2 × 5 – 3, 3 × 5, and 5 × 5 – 2

or, 10 – 3, 15, and 25 – 2

or, 7, 15 and 23

 

∴ Three numbers are: 7, 15, and 23.

 

Example 4: Is -4 a term of the arithmetic sequence, 18, 29/2, 11, … … …

 

Solution:

 

Here,

 

First term (a) = 18

Common difference (d) = 29/2 – 18 = -7/2

n^th term (t_n) = -4

Number of terms (n) = ?

 

By using the formula, we have

 

t_n = a + (n – 1)d

or, -4 = 18 + (n – 1) × -7/2

or, -4 – 18 = -7n/2 + 7/2

or, -22 – 7/2 = -7n/2

or, -51/2 = -7n/2

or, 51 = 7n

or, n = 51/7

 

As the number of terms (n) is a fraction, -4 is not a term of the given series.

 

 

Example 5: Show that, t_n = 4 – 7n is a general term of an A.P. Find its common difference.

 

Solution:

 

Here,

 

t_n = 4 – 7n

 

Therefore,

t₁ = 4 – 7 × 1 = 4 – 7 = -3

t₂ = 4 – 7 × 2 = 4 – 14 = -10

t₃ = 4 – 7 × 3 = 4 – 21 = -17

t₄ = 4 – 7 × 4 = 4 – 28 = -24

 

Hence, it gives the terms

t₁, t₂, t₃, t₄, … … … t_n

i.e. -3, -10, -17, -24, … … …, 4 – 7n

 

Now,

 

Common difference = t₂ – t₁ = -10 – (-3) = -10 + 3 = -7

 

Example 6: If 3^rd term and 9^th term of an A.P. are 20 and 5 respectively, find the 19^th term of the series.

 

Solution:

 

Here,

 

3^rd term (t₃) = 20

9^th term (t₉) = 5

First term (a) = ?

 

By using the formula, we have

 

t_n = a + (n – 1)d

or, t₃ = a + (3 – 1)d

or, 20 = a + 2d …………. (i)

 

Again,

 

t₉ = a + (9 – 1)d

or, 5 = a + 8d …………… (ii)

 

subtracting equation (ii) from (i),

 

20 – 5 = a + 2d – (a + 8d)

or, 15 = a + 2d – a – 8d

or, 15 = -6d

or, d = -15/6 = -5/2

 

Putting the value of d in equation (i),

 

20 = a + 2 × -5/2

or, 20 = a – 5

or, 20 + 5 = a

or, a = 25

 

Hence, the first term is 25 and the common difference is -5/2

 

∴ t₁₉ = a + (n – 1)d

          = 25 + (19 – 1) × -5/2

          = 25 + 18 × -5/2

          = 25 – 45

          = -20

 

Example 7: An arithmetic series is given as 2 + 8 + 14 + 20 + … … … + 80

a. Find the number of terms.

b. Find 8^th term of the series.

 

Solution:

 

Here,

 

First term (a) = 2

Common difference (d) = 8 – 2 = 6

Last term (t_n) = 80

 

a. Number of terms (n) = ?

 

By using the formula,

 

t_n = a + (n – 1)d

or, 80 = 2 + (n – 1)6

or, 80 = 2 + 6n – 6

or, 80 = 6n – 4

or, 80 + 4 = 6n

or, 6n = 84

or, n = 84/6

or, n = 14

 

Hence, the number of terms = 14

 

b. 8^th term (t₈) = ?

 

Again by using the formula,

 

t_n = a + (n – 1)d

or, t₈ = 2 + (8 – 1)6

or, t₈ = 2 + 7 × 6

or, t₈ = 2 + 42

or, t₈ = 44

 

Hence, the 8^th term is 44.

 

 

Example 8: In an arithmetic series t₇/t₁₁ = 11/7. Find the value of t₁₈.

 

Solution:

 

Here,

 

t₇/t₁₁ = 11/7

or, (a + 6d)/(a + 10d) = 11/7

or, 11(a + 10d) = 7(a + 6d)

or, 11a + 110d = 7a + 42d

or, 11a + 110d – 7a – 42d = 0

or, 4a + 68d = 0

or, 4(a + 17d) = 0

or, a + 17d = 0

 

Now,

 

t₁₈ = a + (18 – 1)d

or, t₁₈ = a + 17d

i.e. t₁₈ = 0

 

Example 9: The n^th terms of two AP -19 – 12 – 5 + 2 … … … and 1 + 6 + 11 + … … … are equal then find the value of n.

 

Solution:

 

Here,

 

In 1^st AP: -19 – 12 – 5 + 2 … … …

First term (a) = -19

Common difference (d) = -12 + 19 = 7

Now n^th term = t_n = a + (n – 1)d

or, t_n = -19 + (n – 1)7

or, t_n = -19 + 7n – 7

or, t_n = 7n – 26 ……………. (i)

 

In 2^nd AP: 1 + 6 + 11 + … … …

First term (a) = 1

Common difference (d) = 6 – 1 = 5

Now n^th term = t_n = a + (n – 1)d

or, t_n = 1 + (n – 1)5

or, t_n = 1 + 5n – 5

or, t_n = 5n – 4 ……………… (ii)

 

Equating equation (i) and (ii),

7n – 26 = 5n – 4

or, 7n – 5n = -4 + 26

or, 2n = 22

or, n = 11

 

Example 10: A taxi meter reads Rs. 7.00 at the time of starting and Rs. 9.00 for each additional kilometer. If the distance covered is 12km, find the charge read by the taxi meter.

 

Solution:

 

Here,

 

The sequence of charges read by the taxi meter is Rs.7, Rs.(7 + 9), Rs.(7 + 9 + 9), … … … i.e. Rs.7, Rs.16, Rs.25, … … …

 

First term (a) = 7

Common difference (d) = 9

Number of terms (n) = 13

 

∴ t_n = a + (n – 1)d

or, t₁₃ = 7 + (13 – 1)9

          = 7 + 12 × 9

          = 7 + 108

          = 115

 

∴ The total charge read by the taxi meter = Rs.115.