ex_sol(4)

# ex_sol(4) - Section 11.2 Arithmetic Sequences Instructor Ms...

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Unformatted text preview: Section 11.2: Arithmetic Sequences Instructor: Ms. Hoa Nguyen (nguyenascsisuedu) 1 Arithmetic Sequences An arithmetic sequence {an} satisﬁes the following: 0 a1 . an = an—l + d where 0.1 is the ﬁrst term, d is the common difference. 2 Find the n—th term of an arithmetic sequence Given the ﬁrst term a1 and the common difference d: 01 a2=a1+d a3=a1+2d an=a1+(n—l)d » 7, . -: '7 So, the n—th term of an arithmetic sequence is: in” = a1 + (n — 1)d. ‘ Example 1: [11.2.2aPTIThc 5‘” term of an arithmetic sequence with ﬁrst term a] = 1 and common differenced = ~§ is a‘ ._. ) c\ : ’ i o —1 \ ' " 3, 0 None of these i’, a, ' Q «a? I ‘ V (W‘B Cl 1 0 *" ‘ i‘ i <n=s> 1+ (aw-g) _ 4-512-). [11.2.1bP'1‘1The a“ term of an arithmetic sequence with ﬁrst term at = Example 2: 2 and common difference at = 5% is ‘ 0 None of these —— :- g1...— » g» .ln 94,111“ ‘7“ ’ 0 it? — t" 0 ‘3‘“ t" Q'h 3:591. + ( V‘ " 0 J .63 1g ~ in ,, 3 a = 2 + (WM-lg) /// ‘ \ + 3) Example 3: an aL 73I37 . . . H II ’[11,.2.10PT]The 31“ term of the arithmetic sequence {1 3 l 0, - - -} is 0‘ : A. ‘o —10 = H 6}“ 03. ai + (NO <1 {1 (4 oz : aim 2 :32 n=?>\) ___ (bl-D(—3>‘ :02 ’0‘1 Example4: ’1 It ’- = — a [11.2.2bP'f‘]Gi§en an arithmetic sequence with an = 65, and a“ = 140, ﬁnd the common difference. 0 H4 a : (n'bcl:01‘+‘Q-OC\:C 9‘5“, 2| _ H ,72 -- ~‘ 2 g (214;) : d1 '* («v-Dd 7 aiJerC‘ﬂHO 1|) a. II 4.! \m H O3 ExampleS: (n:uc>) :_._> Q4G_Q2‘: [ll.2.2aPT]Given an arithmetic sequence with am 2 62, and (145 m 13?, ﬁnd the ﬁrst. term a1. "0 3’ CLZI : 01+ =- 62. g: (Me, = (1.1.4166? '3’} o 4 éaﬂﬁo = 25A: “+5 =7 (3:13:33 =5oizg2—2oa 2| , _ , 25 z 9 2—20 5 3 Summlng the terms of arithmetlc sequence 1 Let SkUc = 1, ...,n) be the sum of the ﬁrst 19 terms of an arithmetic sequence {can}. Then 31 = (11 32 = a1+a2=a1+(a1+d)=2a1+d 3 53 = a1+ag+a3=a1+(a1+d)+(a1+2d) =3a1+xni S,1 = a1+az+a3+...+an=a1+(a1+d)+(a1+2d)+...+[a1+(n—1)d] = m1] + [1+ 2 + + (n-— 1)]d = 11111 + n(n2_1)d Note: S = na1+w:‘{712—_—11d = g(2a1+(n — 1)d) = nag—Ga So, E, = nal + ‘ a Examp e : J- a N f/ ‘9" d —; a -0“: 5 [11.2.2cP'flGNen the arithmetic sequence {—4,1,6,11,~-}, ﬁnd the sum 3;”, _ “(n‘-\) 32:38? 390: “01* 01.4 333:3 ‘ (mm 2,390“) + 9033 5- ' 2 2 ~320 + 15900 = lb’hZO ...
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