series_and_sequences assignment 7 - 1. The second term of...

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Unformatted text preview: 1. The second term of an arithmetic sequence is 7. The sum of the first four terms of the arithmetic sequence is 12. Find the first term, a , and the common difference, d , of the sequence. Working: Answer: ………………………………………….. (Total 4 marks) 2. Let a be the first term and d be the common difference, then a + d = 7 and S 4 = 2 4 (2 a + 3 d ) = 12 (M1) = + = + ⇒ 12 6 4 7 d a d a (M1) ⇒ a = 15, d = –8 (A2)(C2)(C2) [4] 3. The ratio of the fifth term to the twelfth term of a sequence in an arithmetic progression is 13 6 . If each term of this sequence is positive, and the product of the first term and the third term is 32, find the sum of the first 100 terms of this sequence. (Total 7 marks) 4. Let the arithmetic sequence be written as a , a + d , a + 2d , ... Then 13 6 11 4 = + + d a d a (M1) So 13 a + 52 d = 6 a + 66 d ⇒ 7 a = 14 d (M1) ⇒ a = 2 d . (A1) Since each term is positive, both a and d are positive. We are given a ( a + 2 d ) = 32, setting a = 2 d , we get 2 d (2 d + 2 d ) = 8 d 2 = 32. (M1) ⇒ d = ± 2. (A1) 1 Hence, d = 2 and a = 4 and sum to 100 terms of this sequence is 2 100 {(2)(4) + (100 – l)2}. (M1) = 10 300 (A1) [7] 5. An arithmetic sequence has 5 and 13 as its first two terms respectively. (a) Write down, in terms of n , an expression for the n th term, a n . (b) Find the number of terms of the sequence which are less than 400. Working: Answers : (a) .................................................................. (b) .................................................................. (Total 4 marks) 6. (a) a 1 = 5 and d = 8 (M1) a n = a 1 + ( n – 1) d a n = 8 n – 3 (A1) (C2) (b) 8 n – 3 < 400 (M1) 8 n < 403 n < 50.375 or n < 50 8 3 or n < 51 Therefore, there are 50 terms less than 400. (A1) (C2) [4] 2 7. The sum of the first n terms of an arithmetic sequence is S n = 3 n 2 – 2 n. Find the n th term u n . Working: Answer: .................................................................. (Total 3 marks) 8. u n = S n – S n –1 (M1) = [3 n 2 – 2 n ] – [3( n – 1) 2 – 2( n – 1)] (A1) = 6 n – 5 (A1) (C3) OR u 1 + u n = 6 n – 4 (M1)(A1) u 1 = 1 ⇒ u n = 6 n – 5 (A1) (C3) [3] 3 9. The probability distribution of a discrete random variable X is given by P ( X = x ) = k x 3 2 , for x = 0, 1, 2, ...... Find the value of k. Working: Answer: .................................................................. (Total 3 marks) 10. Since X is a random variable, ∑ = x x X all ) P( = 1 Therefore, k + 3 2 3 2 3 2 3 2 + + k k k + ......... = 1 (M1) k - 3 2 1 1 = 1 (M1) k = 3 1 (A1) (C3) [3] 4 11. Find the sum of the positive terms of the arithmetic sequence 85, 78, 71, .......
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This note was uploaded on 09/19/2010 for the course MATH 1004 taught by Professor Wilson during the Spring '10 term at International Institute of Information Technology.

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series_and_sequences assignment 7 - 1. The second term of...

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