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Chapter 5: Arithmetic Progressions
Arithmetic Progression
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 = t_{1} = a
Second term = t_{2} = 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
Maths by Mr. M. P. Keshari
