algebraic_methods_only.pdf

# algebraic_methods_only.pdf - Number and Algebra 52 6 A...

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A Number and Algebra 6 Arithmetic and geometric progressions 6.1 Arithmetic progressions When a sequence has a constant difference between successive terms it is called an arithmetic progres- sion (often abbreviated to AP). Examples include: (i) 1, 4, 7, 10, 13, . . . where the common differ- ence is 3 and (ii) a , a + d , a + 2 d , a + 3 d , . . . where the common difference is d . If the first term of an AP is ‘ a ’ and the common difference is ‘ d ’ then the n ’th term is: a + ( n 1) d In example (i) above, the 7th term is given by 1 + (7 1)3 = 19 , which may be readily checked. The sum S of an AP can be obtained by multi- plying the average of all the terms by the number of terms. The average of all the terms = a + l 2 , where ‘ a is the first term and l is the last term, i.e. l = a + ( n 1) d , for n terms. Hence the sum of n terms, S n = n a + l 2 = n 2 { a + [ a + ( n 1) d ] } i.e. S n = n 2 [2 a + ( n 1) d ] For example, the sum of the first 7 terms of the series 1, 4, 7, 10, 13, . . . is given by S 7 = 7 2 [2(1) + (7 1)3], since a = 1 and d = 3 = 7 2 [2 + 18] = 7 2 [20] = 70 6.2 Worked problems on arithmetic progressions Problem 1. Determine (a) the ninth, and (b) the sixteenth term of the series 2, 7, 12, 17, . . . 2, 7, 12, 17, . . . is an arithmetic progression with a common difference, d , of 5. (a) The n ’th term of an AP is given by a + ( n 1) d Since the first term a = 2, d = 5 and n = 9 then the 9th term is: 2 + (9 1)5 = 2 + (8)(5) = 2 + 40 = 42 (b) The 16th term is: 2 + (16 1)5 = 2 + (15)(5) = 2 + 75 = 77 . Problem 2. The 6th term of an AP is 17 and the 13th term is 38. Determine the 19th term. The n ’th term of an AP is a + ( n 1) d The 6th term is: a + 5 d = 17 (1) The 13th term is: a + 12 d = 38 (2) Equation (2) equation (1) gives: 7 d = 21, from which, d = 21 7 = 3. Substituting in equation (1) gives: a + 15 = 17, from which, a = 2. Hence the 19th term is: a + ( n 1) d = 2 + (19 1)3 = 2 + (18)(3) = 2 + 54 = 56 . Problem 3. Determine the number of the term whose value is 22 in the series 2 1 2 , 4, 5 1 2 , 7, . . . 2 1 2 , 4, 5 1 2 , 7, . . . is an AP where a = 2 1 2 and d = 1 1 2 . 52 NUMBER AND ALGEBRA Hence if the n ’th term is 22 then: a + ( n 1) d = 22 i.e. 2 1 2 + ( n 1) ( 1 1 2 ) = 22 ( n 1) ( 1 1 2 ) = 22 2 1 2 = 19 1 2 . n 1 = 19 1 2 1 1 2 = 13 and n = 13 + 1 = 14 i.e. the 14th term of the AP is 22 . Problem 4. Find the sum of the first 12 terms of the series 5, 9, 13, 17, . . . 5, 9, 13, 17, . . . is an AP where a = 5 and d = 4. The sum of n terms of an AP, S n = n 2 [2 a + ( n 1) d ] Hence the sum of the first 12 terms, S 12 = 12 2 [2(5) + (12 1)4] = 6[10 + 44] = 6(54) = 324 Problem 5. Find the sum of the first 21 terms of the series 3.5, 4.1, 4.7, 5.3, . . . 3.5, 4.1, 4.7, 5 . 3, . . . is an AP where a = 3 . 5 and d = 0 . 6. The sum of the first 21 terms, S 21 = 21 2 [2 a + ( n 1) d ] = 21 2 [2(3 . 5) + (21 1)0 . 6] = 21 2 [7 + 12] = 21 2 (19) = 399 2 = 199.5 Now try the following exercise. Exercise 28 Further problems on arith- metic progressions 1. Find the 11th term of the series 8, 14, 20, 26, . . . [68] 2. Find the 17th term of the series 11, 10.7, 10.4, 10.1, . . . [6.2] 3. The seventh term of a series is 29 and the eleventh term is 54. Determine the sixteenth term. [85.25] 4. Find the 15th term of an arithmetic progres- sion of which the first term is 2.5 and the tenth term is 16. [23.5] 5. Determine the number of the term which is 29 in the series 7, 9.2, 11.4, 13 . 6, . . .

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