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Te Hiringa i te Mahara Arithmetic Sequences and Series 2 Kia mahara ake: The nth term in an arithmetic sequence is illustrated by t n a (n 1)d where a is the 1st term in the sequence and d is the fixed difference (common difference) between consecutive terms. The sum for an arithmetic sequence to the nth term is illustrated by S n n2 [2a (n 1)d ] , where a is the 1st term in the sequence and d is the fixed difference (common difference) between consecutive terms; and n is the number of terms. Another equivalent formula is S n n2 [a l ] , where l is the last term. Hei Mahi (I) (a) Find the sum for the first 10 terms in the series 1) Show that the sequence tn:{0.9, 0.7, 0.5, …} is an arithmetic sequence. 2) Determine the first 3 terms of the sequence tn = 2n6, n {1, 2, 3, …}. 3) Find the rule for the arithmetic sequence 1 3 5 tn:{ , , , …}. 8 8 8 4) Find the 10th term of the arithmetic sequence tn:{10, 6, 2, …}. 5) The 3 consecutive terms of an arithmetic sequence are 3.6, y, 8.2. Find the value of y. 6) Insert 3 evenly spaced numbers between -2 and 10. 7) Find the 10th term of the arithmetic sequence where the first term is 5 and the 4th term is 17. 8) Find the difference between the 4th term and the 10th term of the arithmetic sequence tn:{2, 1, 4, …}. 9) The 10th term in an arithmetic sequence is 8 and the 4th term is -4. Determine the first term a. 10) Find the sum to first 8 terms of the arithmetic sequence tn:{14, 12, 10, …}. A project managed by Gardiner and Parata Limited for the Ministry of Education Te Hiringa i te Mahara NGĀ WHAKAUTU Hei Mahi (I) (a) 1) t 2 t1 0.7 0.9 0.2 t 3 t 2 0.5 0.7 0.2 Common difference = 0.2 Thus, it is an arithmetic sequence. 2) t1 2 1 6 4 t 2 2 2 6 2 t3 2 3 6 0 The first 3 terms are 4, 2, and 0. 3) 1 2 1 a ,d 8 8 4 1 1 t n (n 1) 8 4 1 n 1 tn 8 4 4 n 1 tn 4 8 This is the required rule. 4) a 10, d 4 t10 10 (10 1) 4 t10 10 9 4 t10 26 The 10th term is 26. 5) (3.6 8.2) y 2 y 5.9 6) a 2, t 5 10 2 (5 1)d 10 2 4d 10 4d 12 12 d 4 d 3 Thus we have -2, 1, 4, 7, 10 evenly spaced. A project managed by Gardiner and Parata Limited for the Ministry of Education Te Hiringa i te Mahara 7) 8) 5, t 4 17 17 17 12 12 d 3 d 4 Now find t10 : t10 5 (10 1) 4 t10 5 9 4 t10 41 a 2, d 3 t 4 2 (4 1) 3 a 5 (4 1)d 5 3d 3d t4 2 3 3 t 4 7 t10 2 (10 1) 3 t10 2 9 3 t10 25 9) Difference 7 25 18 a (10 1)d t10 a 9d 8 [1] a (4 1)d 4 a 3d 4 [2] Subtract [2] from [1] 6d 12 12 6 d 2 Now replace d in [1] a 9 2 8 d a 18 8 a 8 18 a 10 A project managed by Gardiner and Parata Limited for the Ministry of Education Te Hiringa i te Mahara 10) a 14, d 2 8 S 8 [2 14 (8 1) 2] 2 S 8 4[28 14] S 8 4 14 S 8 56 A project managed by Gardiner and Parata Limited for the Ministry of Education