1)What Is Arithmetic Progression

2) Patterns3) Nth Term4) Sum Of N Terms

In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15 … is an arithmetic progression with common difference of 2.

A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series.

3 , 7 , 11 , 15 , 19 , 23 ……

: Commom Difference (d) = 4 (11 , 15) : Starting Number (a) = 3

If the initial term of an arithmetic progression is and the common difference of successive members is d, then the nth term of the sequence: tn = a +

(n-1)d

Each time you want another term in the sequence you’d add d. This would mean the second term was the first term plus d. The third term is the first term plus d plus d (added twice). The fourth term is the first term plus d plus d plus d (added three times). So you can see to get the nth term we’d take the first term and add d (n - 1) times.

dnaan 1Try this to get the 5th term. 1916341535 a

3 , 7 , 11 , 15 , 19 , 23 ……

1

32

1

32 4

a+d a+2d

a+3d

a

: tn = a + (n-1)d: = 1 + (4-1)4: = 1 + ( 3 ) 4 = 13

a = 1d =

4

a a+d a+2d

a+3d a+4d a+5d

a+6d a+7d a+8d

a = 1d = 3

a a+d a+2d

The fourth term is 3 and the 20th term is 35. Find the first term and both a term generating formula and a recursive formula for this sequence.How many

differences would you add to get from the 4th term to the 20th term? daa 16420 Solve this for dd =

2The fourth term is the first term plus 3 common differences.

daa 314 3 (2)

31 a We have all the info we need to express these sequences.

353 204 aa

35

3

The sum of n terms, we find as,

Sum = n [(first term + last term) / 2]

=Now last term will be = a + (n-1) d

Therefore, Sum(Sn) =n [{a + a + (n-1) d } /2 ]

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

•Solution.

1) First term is a = 100 , an = 500

2) Common difference is d = 105 -100 = 5

3) nth term is an = a + (n-1)d

4) 500 = 100 + (n-1)5

5) 500 - 100 = 5(n – 1)

6) 400 = 5(n – 1)

7) 5(n – 1) = 400

8) 5(n – 1) = 400

9) n – 1 = 400/5

10) n - 1 = 80

11) n = 80 + 1

12) n = 81

Hence the no. of terms are 81.

Problem - Find number of terms of A.P. 100, 105, 110, 115……500