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Example 1: Find the 10th term and the nth term for the sequence 7, 10, 13, … . Solution: a7 d 3 U10 = 7 10 1 3 34 Un= 7 n 1 3 3n 4 1 Example 2 Find the three numbers in an arithmetic progression whose sum is 24 and whose product is 480. Solution Let the three numbers be (a-d), a, (a+d) where d is the common difference. (a-d)(a)(a+d)=480 (a-d) +a + (a+d) = 24 Substituting a =8, 3a = 24 we get a=8 d 2 2 When d = 2, required numbers are 6, 8, 10 When d = -2 , required numbers are 10, 8, 6 3 Example 3 The sum of the first eighteen terms of an arithmetic series is -45 and the eighteenth term is also -45. U 18 45 Find the common difference and the sum of the first hundred terms. Solution Sn n 2 2 a n 1 d S 18 45 18 [ 2 a (18 1) d ] 2 18 a 153 d 45 --------- (1) 4 U 18 45 a 18 1 d 45 a 17 d 45 ------------ (2) Solving equation 1 and 2, we get a 40 d 5 5 Example 4 Find the sum of the positive integers which are less than 500 and are not multiples of 11. Solution Sum of integers Required sum= from 1 to 499 (SI) S I 1 2 3 4 .......... .. 499 499 1 499 2 124750 Sum of multiples of 11(SII) 6 S II 11 22 33 44 ....... 495 11 1 2 3 4 ........ 45 45 1 45 11 2 Sn n a l 2 11385 Required sum = SI-SII =124750-11385 =113365 7 Example 5 S 10 145 The sum of the first 10 terms of an A.P. is 145 and the sum of the next 6 terms is 231 Find (i) the 31st term , and U 31 a 30 d (ii) the least number of terms required for the sum to exceed 2000. S10 U U 11 12 U 13 U 14 U 15 U 16 231 145 S16 16 2 a 15 d 2 8 Solution S16 16 2 2 a 15 d 145 231 82 a 15 d 376 2 a 15 d 47 (1) S 10 145 10 2 2 a 9 d 145 2 a 9 d 29 (2) 9 (1)(2): 6 d 18 d 3 From (2): a 29 9 3 1 2 U 31 a 30 d 1 30 3 91 (ii)Find the least number of terms required for the sum to exceed 2000. Let n be the least number of terms required for S n 2000 10 n 2 n 2 2 a n 1 d 2000 2 1 n 1 3 2000 n 3 n 1 4000 3n 2 n 4000 0 2 Consider 3 n n 4000 0 1 1 4 3 4000 n 36 . 7 or 36 . 3 6 11 For 3n 2 n 4000 0 + n 36 . 3 or n 36 . 7 36 . 3 + 36 . 7 Since n must be a positive integer, n > 36.7 Hence, the least number of terms required is 37. 12 Example 6 Given that the fifth term of a geometric progression is 81 4 and the third term is 9. Find the first term and the common ratio if all the terms in the G.P. are positive. ar 4 81 (1) 4 ar 2 9 (2) 13 (1)(2): ar ar 4 2 r 2 81 4 1 9 9 4 Since all the terms in the G.P. are positive, r>0 3 r 2 2 From (2): a 3 9 2 a4 14 Example 7 Three consecutive terms of a geometric x 1 x progression are 3 , 3 and 81. Find the value of x. If 81 is the fifth term of the geometric progression, find the seventh term. x 1 r 3 3 3 x 3 x 1 3 x 81 3 3 x 1 4 x 1 15 3 x 33 1 3 x 81 r x2 3 x 1 3 4 3 3 3 Given 81 is the fifth term, find the seventh term: U5= ar4 = 81 a 3 81 4 a 81 3 U7 = ar6 1 4 1( 3 ) 729 6 16 Example 8 Find the sum of the first eight terms of the series 3 2 4 8 ......... 3 Soln: 2 a 1 r 1 r 8 U2 8 2 3 1 3 2 1 3 2 U3 2 U4 2 , , a 3, r 3 U1 3 U 2 3 U 3 3 G.P. with S8 9 17 8 2 3 1 3 2 1 3 256 3 1 3 6561 6305 729 18 Example 9 a 1 A geometric series has first term 1 and S5 the common ratio r, where r 1, is positive. The sum of the first five terms is twice the sum of the terms from the 6th to 15th inclusive. Prove that r 5 1 ( 3 1). U 6 U 7 ..... U 15 2 U 1 U 2 U 3 U 4 U 5 U 6 U 7 ........ U 15 U 1 U 2 U 3 U 4 U 5 S15 S 5 19 Solution S 5 2 S15 S 5 Given 3 S 5 2 S 15 a ( r 5 1) a ( r 15 1) 3 2 r 1 r 1 Since r 1 and a = 1, we have 5 3 ( r 1) 2 ( r 2r 15 3r 5 3 5 15 1) 1 0 5 2(r ) 3r 1 0 20 Let x r 5 . So the equation becomes 3 2x 3x 1 0 Let f(x) = 2 x 3 3 x 1 Since f(1) = 2 – 3 + 1 = 0 (x –1) is a factor of f(x). 3 2 x 3 x 1 ( x 1)( 2 x 2 kx 1) Comparing coefficient of x: 3 1 k k 2 3 2 x 3 x 1 ( x 1)( 2 x 2 2 x 1) 21 ( x 1)( 2 x Hence, 2 2 x 1) 0 x 1 or 2 x x 2 2x 1 0 2 4 4 ( 2 )( 1) 2(2) 5 r 1 or r r 1 5 2 12 22 3 4 1 4 ( 3 1) or 2 5 1 ( 3 1) 2 (r<0) 1 Since r 1 and r > 0, r ( 3 1) 2 22 Sum to infinity, S (or S ) = a 1 r The sum to infinity exists (the series converges or series is convergent) provided r 1 Example 10 Determine whether the series given below converge. If they do, give their sum to infinity. G.P. with r 2 >1 (a) 2 4 8 16 ........ Does not converge 23 (b) 1 1 4 16 S 1 1 64 a 1 r .......... . G.P. with r 1 r <1 1 4 4 Series converges 1 1 1 4 4 5 24 (c) 3 9 2 27 4 .......... ..........G.P. 3 with r 2 3 >1 r 2 Does not converge 25 Example 11 A geometric series has first term a and the common ratio 1 . 2 Show that the sum to infinity of the geometric progression is a ( 2 2 ) . Solution S a 1 r a (2 2 1 a 1 1 2 2 ) a (2 a 2 2 1 2) 2 1 2 1 (shown) 26 Example 12 A geometric series has first term a and common ratio r. S is the sum to infinity of the series, T is the sum to infinity of the evennumbered terms (i.e.U 2 U 4 U 6 ....) of the series. Given that S is four times the value 2 common ratio r of T, find the value of r. S a 1 r T ar ar 3 ar 5 ......... ar 1 r2 Given S = 4T, we have a 1 r 4 ar 1 r 2 27 1 1 r 1 4r (1 r )( 1 r ) 4r 1 r 1 r 4r r 1 3 28