# CLASS-10INTRODUCTION & SEQUENCE

ARITHMETIC PROGRESSION (A.P)

Sequence –

Some numbers arranged in a definite order according to some definite rule are said to form a sequence. The number occurring at the nth place of a sequence is called its nth term, denoted by Tₑ or aₑ.

Example.) Consider the rule, Tₑ= (3e + 5)

Putting, e = 1, 2, 3, 4, 5, 6, 7,……………., we get

T₁ = (3e + 5) = 8

T₂ = (3e + 5) = 11

T₃ = (3e + 5) = 14, and so on ……..

Thus the numbers 8, 11, 14, 17, 20, 23,………… form a sequence. In this sequence, the first term is 8, the second term is 11, the third term is 14, and so on. The eth term is (3e + 5). The eth term in a sequence is called its general term.

Arithmetic Progression (A.P) –

A sequence, in which each term differs from its preceding term by a constant, is called an arithmetic progression which can be written as A.P. The constant difference is called the common difference.

Example.1) Show that the progression 7, 12, 17, 22, 27,………..is in A.P. Write its first term and common difference.

Ans.) The given progression is 7, 12, 17, 22, 27,………..

We have, (12 – 7) = 5

(17 – 12) = 5

(22 – 17) = 5

(27 – 22) = 5

So, (12 – 7) = (17 – 12) = (22 – 17) = (27 – 22) = 5 (constant)

So, the given progression is an A.P

It’s first term = 7, and common difference = 5     (Ans.)

Example.2) Show that the progression 7, 4, 1, - 2, - 5, - 8,……….. is n A.P. Write its first term and the common difference

Ans.) The given progression is 7, 4, 1, - 2, - 5, - 8,………..

We have, (4 – 7) = - 3,

(1 – 4) = - 3,

(- 2 – 1) = - 3,

{- 5 – (- 2)} = (- 5 + 2) = - 3

{- 8 – (- 5)} = (- 8 + 5) = - 3

So, (4 – 7) = (1 – 4) = (- 2 – 1) = {- 5 – (- 2)} = {- 8 – (- 5)} = - 3 (constant)

So, the given progression is an A.P.

Its first term = 7, and common difference = - 3. (Ans.)

Example.3) Show that each of the following progressions is an A.P. Find the common difference and the next term of each –

(i) √7, √28, √63,…………

Ans.) The given terms are √7, √28, √63,…………

We write them as √7, √(4 X 7), √(9 X 7),…………

=> √7, √(2² X 7), √(3² X 7),…………

=> √7, 2√7, 3√7,…………

We have, (2√7 - √7) = √7

(3√7 - 2√7) = √7

So, (2√7 - √7) = (3√7 - 2√7) = √7       (constant)

So the given progression is an A.P. in which the common difference is √7. Clearly the next term is 4√7 = √(4² X 7) = √(16 X 7) = √112     (Ans.)

(ii) √18, √50, √98,…………….

The given terms are, √18, √50, √98,…………….

We may write as, √(9 X 2), √(25 X 2), √(49 X 2),…………….

=> √(3² X 2), √(5² X 2), √(7² X 2),…………..

=> 3√2, 5√2, 7√2,……………

We have, (5√2 - 3√2) = 2√2

(7√2 - 5√2) = 2√2

So, (5√2 - 3√2) = (7√2 - 5√2) = 2√2      (constant)

So the given progression is an A.P. in which the common difference is 2√2. Clearly the next term is 9√2 = √(9² X 2) = √(81 X 2) = √162     (Ans.)