# 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, … … …, tn be the first, second, third, fourth, … … …, nth 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

… … …

tn = 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 7th 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

7th term (t7) = ?

We have,

tn = a + (n – 1)d

t7 = 7 + (7 – 1)3

= 7 + 6 × 3

= 7 + 18

= 25

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

Solution:

Here,

15th term (t15) = 92

First term (a) = 8

Common difference (d) = ?

By using the formula, we have

tn = 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

nth term (tn) = -4

Number of terms (n) = ?

By using the formula, we have

tn = 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, tn = 4 – 7n is a general term of an A.P. Find its common difference.

Solution:

Here,

tn = 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, … … … tn

i.e. -3, -10, -17, -24, … … …, 4 – 7n

Now,

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

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

Solution:

Here,

3rd term (t3) = 20

9th term (t9) = 5

First term (a) = ?

By using the formula, we have

tn = 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 8th term of the series.

Solution:

Here,

First term (a) = 2

Common difference (d) = 8 – 2 = 6

Last term (tn) = 80

a. Number of terms (n) = ?

By using the formula,

tn = 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. 8th term (t8) = ?

Again by using the formula,

tn = a + (n – 1)d

or, t8 = 2 + (8 – 1)6

or, t8 = 2 + 7 × 6

or, t8 = 2 + 42

or, t8 = 44

Hence, the 8th 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 nth terms of two AP -19 – 12 – 5 + 2 … … … and 1 + 6 + 11 + … … … are equal then find the value of n.

Solution:

Here,

In 1st AP: -19 – 12 – 5 + 2 … … …

First term (a) = -19

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

Now nth term = tn = a + (n – 1)d

or, tn = -19 + (n – 1)7

or, tn = -19 + 7n – 7

or, tn = 7n – 26 ……………. (i)

In 2nd AP: 1 + 6 + 11 + … … …

First term (a) = 1

Common difference (d) = 6 – 1 = 5

Now nth term = tn = a + (n – 1)d

or, tn = 1 + (n – 1)5

or, tn = 1 + 5n – 5

or, tn = 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

tn = 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.