SERIES AND SEQUENCE. solution. 3r + 2r 2. r=1. r=1 = = = 155. solution. r 3. 2r +

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1 Series SERIES AND SEQUENCE Sequence 1,, 3, 4,... example The series n = n r=1 r n r=1 r = n = n(n + 1) Eg.1 The sum of the first 100 natural numbers is given by n=100 r=1 r = 100(101) = 5050 Eg. Find the sum of the series 5 r=1 r(3 + r) 5 r(3 + r) = r=1 5 (3r + r ) r=1 = 5 5 3r + r r=1 r=1 = = = 155 Eg.3 Find the sum of the series 4 r=1 (r + r3 ) 4 (r + r 3 ) = r=1 4 r + 4 r=1 r=1 r 3 = = = 10 1

2 EXX Evaluate the following: i. The sum of the first n natural numbers = n(n + 1) ii. The sum of the squares of the first n natural numbers = iii. The sum of the cubes of the first n natural numbers [ SEQUENCE n(n + 1)(n + 1) 6 n(n + 1) ] A sequence is a set of quantities u 1, u,..., stated in a definite order and each term formed according to a fixed pattern. Example 1, 3, 5, 7,.., n, 6, 18, 54,... FINITE SEQUENCE A finite sequence contains only a finite number of terms. INFINITE SEQUENCE An infinite sequence is unending SERIES A series is formed by the sum of the terms of the sequence. ARITHMETIC SEQUENCE This is a sequence in which the difference between any two successive terms is some constant d. For any arithmetic sequence, u n+1 = u n + d d = u n+1 u n The first term of an arithmetic sequence is denoted by a The general term or nth term of an arithmetic sequence is given by u n = a + (n 1)d where

3 a = 1st term d = common difference n =number of terms The general arithmetic series can be written as a + (a + d) + (a + d) + (a + 3d) +... where a = 1st term d = common difference The nth term = a + (n 1)d and the sum to n term, S n is given by S n = n [a + (n 1)d] 1. Find the 15th term of the sequence, 5, 8,... Example a =, d = (5 ) = (8 5) = 3 u n = a + (n 1)d u 15 = + (15 1)(3) u 15 = + 14(3) u 15 = 44. The 7th term of an AP is 3 and the 1th term is 3. Find (i) the first term and the common difference (ii) the 10th term 3

4 (i) u 7 = a + 6d = 3...(1) u 1 = a + 11d = 3...() eqn(1) () 5d = 6 d = 6 5 a + 6( 6 5 ) = 3 a + ( 36 5 ) = 3 5a 36 = 15 5a = 51 a = 51 5 (ii) u n = a + (n 1)d u 10 = (10 1)( 6 5 ) = ( 6 5 ) = = Find the value of x if x 1, x and 3x + 4 are consecutive terms of an AP. 4

5 Here we apply the principle that; u n+1 u n = d u u 1 = u 3 u = d (x ) (x 1) = (3x + 4) (x ) x x + 1 = 3x + 4 x + x 1 = x + 6 3x = 7 x = If the 7th term of an AP is and the 1th term is 37, find the series. a + 6d =...(1) a + 11d = 37...() () (1) 5d = 15 d = 3 from(1) a + 6(3) = a + 18 = a = 14 The series is given by

6 ARITHMETIC MEAN The arithmetic mean of x and y is given by x + y The arithmetic mean of two numbers is simply their average. 1. Insert five numbers between 5 and 17. EXAMPLE This means that we need to find 7 terms within u 1 = 5 and u 7 = 17 u 1 = a = 5 u 7 = 5 + 6d = 17 6d = 1 d = The numbers are 5, 7, 9, 11, 13, 15, 17. Insert 3 arithmetic means between 8 and 18. Let the means be x, y and z 8 + x + y + z + 18 form an AP 1st term = 8 = a 5th term = 18 = a + 4d 18 = 8 + 4d d = 5 x = = 1 y = = 13 z = = 31 8, 1, 13, 31, 18 form an AP. 6

7 1. Find 5 arithmetic means between 1 and 1.6. Insert 5 arithmetic means between 3 and 15 EXERCISE 7

8 THE SUM,S n OF AN ARITHMETIC SEQUENCE S n = a + [a + d] a + (n )d + a + (n 1)d...(1) Reversing S n = a + (n 1)d + [a + (n )d] [a + d] + a...() Adding (1) and (): S n = n[a + (n 1)d] S n = n [a + (n 1)d] Example 1. Find the sum of the first twenty terms of an AP 1,4,7,10,... a = 1, d = 3, n = 0 Solution S n = n [a + (n 1)d] = 0 [(1) + (0 1)(3)] = 10[ + 19(3)] = 10( + 57) = 590. If the 7th term of an AP is and the 1th term is 37, find the sum of the first 10 terms. Solution 7th term = a + 6d =...(1) 1th term = a + 11d = 37...() () (1) 5d = 15 d = 3 F rom(1) a + 6(3) = a + 18 = a = 4 The sum to n terms is given by S n = n [a + (n 1)d] S 10 = 10 [(4) + (10 1)(3)] = 5[8 + 9(3)] = 5[8 + 7] = 175 8

9 3. The 6th term of an AP is 5 and the 10th term is 1. Find the sum of the first 30 terms. Solution 6th term = a + 5d = 5...(1) 10th term = a + 9d = 1...() () (1) 4d = 16 d = 4 F rom(1) a + 5( 4) = 5 a 0 = 5 a = 15 The sum to n terms is given by S n = n [a + (n 1)d] S 30 = 30 [(15) + (30 1)( 4)] = 15[30 + 9( 4)] = 15[30 116] = 15( 86) Exercise = The 7th term of an AP is 3 and the 1th term is 3. Find the first term and the common difference and the sum to 45 terms.. The 9th term of an AP is 1 and the 15 term is 45. Find the sum to 6 terms. 9

10 GEOMETRIC SEQUENCE (G.S) This is a sequence in which the ratio,r, of any two successive terms is a constant. For any G.P, r = u n+1 u n u n+1 = ru n A G.P has the form a + ar + ar + ar where a =first term and r =common ratio The nth term or the general term of a GP is given by; u n = ar n 1 where a = first term and r =common ratio. Example 1. Find the common ratio of the sequence; 5, 10, 0, 40,... Solution r = 10 5 = 0 10 = 40 0 =. The 5th term of a GP is 16 and the 8th term is 4374, find the common ratio and the first term. 5th term = ar 4 = 16...(1) 8th term = ar 7 = () () (1) : ar 7 ar = r 3 = 7 r = 3 from (1), a(3) 4 = 16 a = = The series is

11 GEOMETRIC MEANS If x,y and z form a geometric sequence, then y = z, where x,y,z 0 x y y = xz y = ± xz, where xz > 0 Thus, the geometric mean, y of two positive numbers x and z is given by y = xz 1. Insert 4 geometric means between 5 and 115 Example Solution Let the numbers be A,B,C,D 5,A,B,C,D,115 form a GP a = 5, ar 5 = 115 r 5 = = 43 r = 3 A = 5 3 = 15 B = 5 3 = 45 The required Geometric series is 5, 15, 45, 135, 405, 115 C = = 135 D = = 405. Insert two geometric means between 5 and

12 Solution 1. The geometric sequence 1, 3, 9 4 Let the means be A and B 5, A, B, 8.64 form a GP a = 5, ar 3 = 8.64 EXERCISE r 3 = 1.78 r = 1. A = 5 1. = 6.0 B = 5 (1.) = 7. 43,... has a term equal to. Find the number of terms. 3 Solution a = 1, r = 3, u n = 43 3 u n = ar n 1 = ( 3 )n 1 = 43 3 ( 3 )n 1 = ( 3 )5 n 1 = 5 n = 6. For which values of k are k, 5k +, and 0k 4 consecutive terms of a geometric sequence? 1

13 Here we apply the principle u n+1 u n = r r = u u 1 = u 3 u u = u 1 u 3 (5k + ) = k(0k 4) 5k + 0k + 4 = 40k 8k 15k 8k 4 = 0 (15k + )(k ) = 0 15k + = 0 or k = 0 k = 15 or k = when k = 4, k = 15 15, 5k + = 4 0, and 0k 4 = 3 3 when k =, k = 4, 5k + = 1, and 0k 4 = 36 {k : k = 15, } 13

14 a, ar, ar, ar 3,... SUM TO n TERMS OF A GEOMETRIC SEQUENCE S n = a + ar + ar ar n 1 rs n = ar + ar + ar 3 + ar n 1 + ar n S n rs n = a ar n S n (1 r) = a ar n = a(1 r n ) S n = a(1 rn ) 1 r The formula for S n is useful when r is a fraction between +1 and 1 but for values of r outside this range, the alternative form S n = a(rn 1) r 1 could be used. Example 1. Find the sum of the first 8 terms of the series Solution a = 8, r = 4 8 = 4 = 1, n = 8 S n = a(1 rn ) 1 r = 8[1 ( 1 )8 ] 1 1 = 8( ) 1 = Find the sum to 6 terms of the G.P 1, 1, 1,... 4 a = 1 4, r = 1 1 = = S n = a(rn 1) r 1 since r > 1 S 6 = 1 4 (6 1) 1 14 = 1 63 (63) = 4 4

15 3. Find the sum of the first 10 terms of the series a = 1, r = 3 4, n = 10 S n = a(1 rn ) 1 r S 10 = 1(1 ( 3 4 )10 ) 1 ( 3 4 ) = 1(1 ( 3 4 )10 ) = Find the sum of the series to 6 terms. = 6.47 a =, r =, n = 6 S n = a(rn 1) r 1, r > 1 S 6 = (6 1) 1 = (3 1) 1 = 6 5. Find the sum to 7 terms of the series

16 a = 3, n = 7, r = 6 3 = 1 6 = 4 1 = S n = a(rn 1) r 1, r < 1 S 7 = 3(( )7 1) 1 = 3( 18 1) 3 = 3( 19) 3 = 19 EXERCISE 1. The 5th, 9thand16th terms of a linear sequence are consecutive terms of an exponential sequence GP. Find the common difference in terms of the first term.. The 3rd and 6th terms of an exponential sequence are 1 and 1 4 first 10 terms. 3 respectively. Find the sum of the 16

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