Arithmetic Progression


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Arithmetic Progression

  1. 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. 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. 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. 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. 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. 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. 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. 8. DEEPALI TANWAR TH – B 10 6