Chapter 5: Arithmetic Progressions

Let us consider the following set of members :
1, 5, 9, 13, 17………….
9, 12, 15, 18, 21 ……..
-5, 0, 5, 10, 15………
1.3, 1.6, 1.9, 2.2 ……..…

All these sets follow certain rules. In first set 5 - 1 = 9 - 5 = 13 - 9 = 17 - 13 = 4

In second set 12 - 9 = 15 - 12 = 18 - 15 = 21 - 18 = 3

and so on. In first set the number after 17 in 17 + 4 = 21 and in second set number after 21 is 21 + 3 = 24. In this way we find that in first set second number is 1+ 4 = 5, third number is 5 + 4 = 9 = 1 + 2 x 4 and son on.

On the basis of above discussion we can consider the following series

a, a + d, a + 2d, a + 3d, .........................

Here a = 1, d = 4

a + d = 1 + 4 = 5

a + 2d = 1 + 2 x 4 = 9

and so on

Thus we can say that

a = First term

a + d = Second term

a + 2d = Third term

a + 3d = Fourth term and son on

n^th term = a + (n - 1)d

Here First term = t₁ = a

Second term = t₂ = a + d

and hence, t_n = a + (n - 1)d

d is called common difference and the series is called arithmultic progression

Let s = 1 + 2 + 3 + 4 + 5 + 6 + 7 + + 8 + 9 + 10 and writting in reversed order

S = 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1

Adding these two we get 2s = 11 + 11 + 11 + 11 + 11 + 11 + 11 = 11 + 11 + 11

= 10 X 11

s = (10 X 11)/2 = 55

In similar way, if

Adding we get