2018-II-12 Posted on 16-06-2021 By app.cch No Comments on 2018-II-12 Ans: AConsider ①a3=a1+a221=a1+a2a1=21–a2 …① Consider ②a4=a2+a3a4=a2+21a2=a4–21 …② Sub. ②② into ①①, we have ③a1=21–(a4–21)a1=42–a4 …③ Consider ④a5=a3+a4a5=21+a4a5–a4=21 …④ Consider ⑤a6=a4+a5a4+a5=89 …⑤ ⑤④⑤–④, we have 2a4=68a4=34 Sub. a4=34 into ③③, we have a1=42–34a1=8 Same Topic: 2018-I-16 2018-II-35 2018-II-43 2018-II-45 2018, HKDSE-MATH, Paper 2 Tags:Sequences