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# Arithmetic Progression

Let a and d be real numbers. Then the numbers of the form a, a + d , a + 2d , a + 3d , a + 4d , ... is said to form Arithmetic Progression denoted by A.P. The number ‘a’ is called the first term and ‘d’ is called the common difference.

Arithmetic Progression

Let us begin with the following two illustrations

### Illustration 1

Make the following figures using match sticks (i) How many match sticks are required for each figure? 3,5,7 and 9.

(ii) Can we find the difference between the successive numbers?

5 − 3 = 7 − 5 = 9 − 7 = 2

Therefore, the difference between successive numbers is always 2.

### Illustration 2

A man got a job whose initial monthly salary is fixed at ₹10,000 with an annual increment of ₹2000. His salary during 1st , 2nd and 3rd years will be ₹ 10000, ₹ 12000 and ₹ 14000 respectively.

If we now calculate the difference of the salaries for the successive years, we get 12000 – 10000 = 2000; 14000 – 12000 = 2000 . Thus the difference between the successive numbers (salaries) is always 2000.

Did you observe the common property behind these two illustrations? In these two examples, the difference between successive terms always remains constant. Moreover, each term is obtained by adding a fixed number (2 and 2000 in illustrations 1 and 2 presented above) to the preceding term except the first term. This fixed number which is a constant for the differences between successive terms is called the “common difference”.

Definition

Let a and d be real numbers. Then the numbers of the form a, a + d , a + 2d , a + 3d , a + 4d , ... is said to form Arithmetic Progression denoted by A.P. The number ‘a’ is called the first term and ‘d’ is called the common difference.

Simply, an Arithmetic Progression is a sequence whose successive terms differ by a constant number. Thus, for example, the set of even positive integers 2, 4, 6, 8, 10, 12,… is an A.P. whose first term is a = 2 and common difference is also d = 2 since 4 − 2 = 2, 6 − 4 = 2, 8 − 6 = 2, …

Most of common real−life situations often produce numbers in A.P.

Note

·           The difference between any two consecutive terms of an A.P. is always constant. That constant value is called the common difference.

·           If there are finite numbers of terms in an A.P. then it is called Finite Arithmetic Progression. If there are infinitely many terms in an A.P. then it is called Infinite Arithmetic Progression.

## Terms and Common Difference of an A.P.

1. The terms of an A.P. can be written as

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 , . . .

In general, the nth term denoted by tn can be written as tn  = a + (n −1)d.

2. In general to find the common difference of an A.P. we should subtract first term from the second term, second from the third and so on.

For example, t1  = a, t 2  = a +d

Therefore, t 2t1  = (a + d ) −a = d

Similarly, t 2  = a +d, t 3 = a + 2d,...

Therefore, t 3t 2  = (a + 2d) −(a + d ) = d

In general, d = t 2t1 = t 3t 2  = t 4t3  = ....

So, d = tn tn1   for n = 2, 3, 4,...

Let us try to find the common differences of the following A.P.’s

(i) 1, 4, 7, 10,….

d= 4 − 1 = 7 − 4 = 10 − 7 = ... = 3

(ii) 6, 2, − 2, −6,…

d = 2 6 = − 2 2 = −6 (−2) = ... = −4

Thinking Corner: If tn is the nth term of an A.P. then the value of tn +1tn −1 is _______.

### Example 2.23

Check whether the following sequences are in A.P. or not?

(i) x + 2,  2x + 3,  3x + 4,….

(ii) 2, 4, 8, 16,...

(iii) 3√2, 5√2, 7√2, 9√2,...

### Solution

To check that the given sequence is in A.P., it is enough to check if the differences between the consecutive terms are equal or not.

(i) t 2 -t1 = (2x + 3) (x + 2) = x + 1

t 3 -t2 = (3x + 4) (2x + 3) = x + 1

t 2 - t1  = t 3 t2

Thus, the differences between consecutive terms are equal.

Hence the sequence x + 2, 2x + 3, 3x + 4,... is in A.P.

(ii) t 2 -t1  = 4 2 = 2

t 3 -t2   = 8 4 = 4

t 2 -t1  = t 3 t2

Thus, the differences between consecutive terms are not equal. Hence the terms of the sequence 2, 4, 8, 16, . . . are not in A.P.

(iii) t2 -t1  = 5√2 − 3√ 2 = 2√ 2

t3 -t2  = 7√2 − 5√ 2 = 2√ 2

t4 -t3  = 9√ 2 − 7√2 = 2√ 2

Thus, the differences between consecutive terms are equal. Hence the terms of the sequence 32, 52, 72, 92,... are in A.P

### Example 2.24

Write an A.P. whose first term is 20 and common difference is 8.

### Solution

First term = a = 20 ; common difference = d= 8

Arithmetic Progression is a, a + d, a + 2da + 3d,...

In this case, we get 20, 20 + 8, 20 + 2(8),  20 + 3(8),...

So, the required A.P. is 20, 28, 36, 44,…..

Note

An Arithmetic progression having a common difference of zero is called a constant arithmetic progression.

### Example 2.25

Find the 15th, 24th and nth term (general term) of an A.P. given by 3, 15, 27, 39,…….

### Solution

We have, first term = a = 3 and common difference = d = 15 3 = 12 .

We know that nth  term (general term) of an A.P. with first term a and common difference d is given by

tn = a + (n −1)d

t15   = a + (15 −1)d = a + 14d = 3 + 14 (12) = 171

(Here a = 3 and d = 12)

t24= a + (24 −1)d = a + 23d = 3 +23(12) = 279

The nth (general term) term is given by

tn = a + (n − 1)d

Thus,  tn = 3 + (n −1)12

tn = 12n − 9

Note

In a finite A.P. whose first term is a and last term l, then the number of terms in the A.P. is given by l = a  + (n −1)d gives n = ( [l-a] / d) +1.

### Example 2.26

Find the number of terms in the A.P. 3, 6, 9, 12,…, 111.

### Solution

First term a = 3 ; common difference d = 6 − 3 = 3 ; last term l = 111 Thus the A.P. contain 37 terms.

### Example 2.27

Determine the general term of an A.P. whose 7th term is −1 and 16th term is 17.

### Solution

Let the A.P. be t1 , t 2 ,t 3 , t 4,...

It is given that  t7 = −1 and t16 = 17

a + (7 −1)d = −1 and a + (16 −1)d = 17

a + 6d = −1 ...(1)

a + 15d = 17 ...(2)

Subtracting equation (1) from equation (2), we get 9d = 18 gives d = 2

Putting d = 2 in equation (1), we get a + 12 = −1  so a = –13

Hence, general term tn  = a + (n −1)d

= −13 + (n −1)×2 = 2n −15

### Example 2.28

If lth , mth  and nth terms of an A.P. are x, y, z respectively, then show that

(i) x (mn ) + y (nl ) + z (lm) = 0 (ii) (xy )n + (yz )l + (z− x )m = 0

### Solution

(i) Let a be the first term and d be the common difference. It is given that

tl  = x, tm  = y, tn  = z

Using the general term formula

a + (l −1)d = x ...(1)

a + (m −1)d = y ...(2)

a + (n −1)d = z ...(3)

We have, x (mn ) + y(nl ) + z (lm)

=a [(m n ) + (n l ) + (l m )] + d [(m n)(l 1) + (n l )(m −1) + (l m)(n −1)]

=a  +d[lm ln m + n + mn lm n + l +ln mn l + m]

=a(0) + d(0) = 0

(ii) On subtracting equation (2) from equation (1), equation (3) from equation (2) and equation (1) from equation (3), we get

x y = (l m)d

y− z = (mn)d

z x = (n l)d

(xy )n + (yz )l + (zx)m = [(lm)n + (mn )l + (nl)m ]d = lnmn + lmnl + nmlm d = 0

Note

In an Arithmetic Progression

·           If every term is added or subtracted by a constant, then the resulting sequence is also an A.P.

·           If every term is multiplied or divided by a non-zero number, then the resulting sequence is also an A.P.

·           If the sum of three consecutive terms of an A.P. is given, then they can be taken as a -d, a and a + d . Here the common difference is d.

·           If the sum of four consecutive terms of an A.P. is given then, they can be taken as a - 3d , a -d , a + d and a + 3d . Here common difference is 2d.

### Example 2.29

In an A.P., sum of four consecutive terms is 28 and their sum of their squares is 276. Find the four numbers.

### Solution

Let us take the four terms in the form (a - 3d), (a -d), (a + d) and (a + 3d) .

Since sum of the four terms is 28,

a 3d +a d +a + d +a + 3d = 28

4a = 28 gives a = 7

Similarly, since sum of their squares is 276,

(a − 3d)2  + (a − d )2  + (a + d )2  + (a + 3d)2 = 276.

a2 − 6ad + 9d2 + a2 − 2ad + d2 +a2 + 2ad +d2 + a2 + 6ad + 9d2  = 276

4a2  + 20d2  =276 4(7)2  + 20d2  = 276.

d2  = 4  gives d = ± 2

If d = 2 then the four numbers are 7 - 3(2),   7 – 2, 7 + 2, 7+3(2)

That is the four numbers are 1, 5, 9 and 13.

If a = 7, d = −2 then the four numbers are 13, 9, 5 and 1

Therefore, the four consecutive terms of the A.P. are 1, 5, 9 and 13.

Condition for three numbers to be in A.P.

If a, b, c are in A.P. then b = a +d, c = a + 2d

so a + c = 2a + 2d

2(a + d) = 2b

Thus 2b

2b = a +c

Similarly, if 2b = a +c, then b − a = c − b so a, b, c are in A.P.

Thus three non-zero numbers a, b, c are in A.P. if and only if 2b = a +c

### Example 2.30

A mother devides ₹207 into three parts such that the amount are in A.P. and gives it to her three children. The product of the two least amounts that the children had ₹4623. Find the amount received by each child.

### Solution

Let the amount received by the three children be in the form of A.P. is given by

a d , a, a + d . Since, sum of the amount is ₹207, we have

(ad ) +a + (a +d) = 207

3a = 207 gives a = 69

It is given that product of the two least amounts is 4623.

(ad )a = 4623

(69 − d )69 = 4623

d = 2

Therefore, amount given by the mother to her three children are

₹(69−2), ₹69, ₹(69+2). That is, ₹67, ₹69 and ₹71.

Tags : Definition, Theorem, Illustration, Terms, Common Difference, Example, Solution | Mathematics , 10th Mathematics : UNIT 2 : Numbers and Sequences
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10th Mathematics : UNIT 2 : Numbers and Sequences : Arithmetic Progression | Definition, Theorem, Illustration, Terms, Common Difference, Example, Solution | Mathematics