# arithmetic progression - \$@mee

Post on 05-Aug-2015

53 views

Category:

## Design

2 download

Embed Size (px)

TRANSCRIPT

1. In mathematics, an arithmetic progression (AP) or arithmeticsequence is a sequence of numbers such that the differencebetween the consecutive terms is constant. For instance, thesequence 5, 7, 9, 11, 13, 15 is an arithmetic progression withcommon difference of 2. 2. A finite portion of an arithmetic progression is called a finitearithmetic progression and sometimes just called an arithmeticprogression. The sum of a finite arithmetic progression is calledan arithmetic series. 3. 3 , 7 , 11 , 15 , 19 , 23 : Commom Difference (d) = 4 (11 , 15): Starting Number (a) = 3 4. If the initial term of an arithmetic progression is and thecommon difference of successive members isd, thenthenth term of the sequence: 5. 3 , 7 , 11 , 15 , 19 , 23 Each time you want another term in the sequence youd add d.This would mean the second term was the first term plus d. Thethird term is the first term plus d plus d (added twice). Thefourth term is the first term plus d plus d plus d (added threetimes). So you can see to get the nth term wed take the first termand add d (n - 1) times.a a n d n 1Try this to get the 5th term.3 5 14 3 16 19 5 a 6. 1 2 3 7. a1 2 3 4 8. a = 1d = 3 9. The fourth term is 3 and the 20th term is 35. Find the first term and both aterm generating formula and a recursive formula for this sequence.How many differences would you addto get from the 4th term to the 20thterm?3 35 4 20 a a a a 16d 20 4 Solve this for d d = 2The fourth term is the first term plus 3common differences.a a 3d 4 1 3 (2)35 33 1 a We have all the info we need to express these sequences. 10. The sum ofn terms, we find as,Sum = n [(first term + last term) / 2]=Now last term will be = a + (n-1) dTherefore,Sum(Sn) =n [{a + a + (n-1) d } /2 ]= n/2 [ 2a + (n+1)d] 11. Solution.1) First term is a = 100 , an = 5002) Common difference is d = 105 -100 = 53) nth term is an = a + (n-1)d4) 500 = 100 + (n-1)55) 500 - 100 = 5(n 1)6) 400 = 5(n 1)7) 5(n 1) = 4008) 5(n 1) = 4009) n 1 = 400/510) n - 1 = 8011) n = 80 + 112) n = 81Hence the no. of terms are 81.