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 CBSE Maths eBooks CBSE Guess > eBooks > Class X > Maths by Mr. M. P. Keshari 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 nth term = a + (n - 1)d Here First term = t1 = a Second term = t2 = a + d and hence, tn = 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