ARITHMETIC PROGRESSION

CHAPTER – 5

SUBMITTED TO = MRS. RENU WADHAWAN

CONTENTDefinations an = a+(n-1)dExamplesSum Of First n Terms Of An APDerivation Of Formula

Sn

DEFINATIONSArithmetic progression(AP) = An AP is a list of numbers in which each term is obtained by adding a fixed numbers to preceding term except the first term. Remember that it can be positive, negative, or zero.

E.G = Reena applied for a job and got selected. She has been offered a job with a starting monthly salary of RS8000, with annual increment of RS500 in her salary. Her salary for the 1st,2nd,3rd,….years will be respectively

8000,8500,9000,9500,…………..

Common Difference = The fixed numbers is called common difference, it is denoted as “d”. a2 - a1 = a3 - a2 = …. = an-

an-1 = d

Sequence = A sequence is an arrangement of numbers in a definite order following definite rule.

Finite Sequence = A sequence containing definite numbers of terms.

Infinite Sequence = A sequence is called infinite sequence if it is not a finite sequence.

General Form Of An AP = a,a+d,a+2d,a+3d represents an AP where a is the first term and d is the common difference. This is called general form of an AP.

DERIVATION OF FORMULA FOR nth TERM OF AN APLet us consider the general form of an AP

a,a+d,a+2d,a+3d…….nth term

This formula can be used it find nth term of an AP whose first term is a and common difference is d

an=a+(n-1)d

1st term = a1 a+(1-1)d = a+0d

a

2nd term = a2 a+(2-1)d = a+1d

a+d

3rd term = a3 a+(3-1)d = a+ 2d

a+2d

4th term = a4 a+(4-1)d = a+3d

a+3d

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An term ( last term)

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

Let us consider some examples

1) Find the 10th term of an AP:2,7,12,…?Sol:- here a=2, d=a2-a1 = 7-2=5, n=10 an=a+(n-1)d a10=2+(10-1)5 =2+(9)5 = 2+45 a10 = 47

2)How many two digit no. are divisible by 3?Sol:- AP: 12,15,18……99Here a=12, d=15-12=3, an= 99 an=a+(n-1)d 99=12+(n-1)3 87=(n-1)3 n-1 = 87/3 = 29

n=29+1 n=29

SUM OF FIRST n TERMS OF AN APLet a be the first term and d be the common difference of an AP .

Let sn denote the sum of first n terms and l be its last term.

Sn = 1+2+3+4+5

Sn = 5+4+3+2+1

2Sn = 30

Sn = 30/2 = 15

Sn = 15

2Sn = 6+6+6+6+6

DERIVATION OF FORMULAa,a+d,a+2d……..

The nth term of this AP is a a+(n-1)d. let s denote the sum

of first n terms of the AP.S=a+(a+d)+(a+2d)+…+[a+(n-1)d]

Rewriting the terms in reversing order,

2s=[2a+(n-1)d]+[2a+(n-1)d]+…+[2a+(n-1)d]+[2a+(n-1)d]

n times

2S=n[2a+(n-1)d]

S=n/2[2a+(n-1)d]

The Sum Of The First n Terms Of An AP Is Given ByS=n/2[2a+(n-1)d]S=n/2[a+a+(n-1)d]S=n/2(a+an)If there is only n terms in AP then an=l, the last term from (3)S=n/2(a+l)

DEEPALI TANWAR

10TH – B

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