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- 1. DEFINATIONS Arithmetic 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
- 2. 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.
- 3. DERIVATION OF FORMULA FOR nth TERM OF AN AP Let us consider the general form of an AP a,a+d,a+2d,a+3d…….nth term 1st term = a1 a+(1-1)d = a+0d 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 --------An term ( last term) a ------------a+(n-1)d = --------------a+(n-1)d 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
- 4. Let us consider some examples 1) Find the 10th term of an AP:2,7,12,…? Sol:- here a=2, d=a2-a1 = 72=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……99 Here 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
- 5. SUM OF FIRST n TERMS OF AN AP Let 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 = 6+6+6+6+6 2Sn = 30 Sn = 30/2 = 15 Sn = 15
- 6. DERIVATION OF FORMULA a,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]
- 7. The Sum Of The First n Terms Of An AP Is Given By S=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)
- 8. DEEPALI TANWAR TH – B 10 6

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