[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-138393-105":59,"doc-detail-138393-en":130},{"code":4,"msg":5,"data":6},0,"success",[7,13,18,23,28,33,38,43,48,51,55],{"id":8,"doc_module":4,"doc_module_name":9,"category_name":10,"show_sort_weight":11,"slug":12},1,"Document","Story & Novel",90,"story-novel",{"id":14,"doc_module":4,"doc_module_name":9,"category_name":15,"show_sort_weight":16,"slug":17},2,"Literature",80,"literature",{"id":19,"doc_module":4,"doc_module_name":9,"category_name":20,"show_sort_weight":21,"slug":22},4,"Exam",70,"exam",{"id":24,"doc_module":4,"doc_module_name":9,"category_name":25,"show_sort_weight":26,"slug":27},5,"Comic",60,"comic",{"id":29,"doc_module":4,"doc_module_name":9,"category_name":30,"show_sort_weight":31,"slug":32},6,"Technology",50,"technology",{"id":34,"doc_module":4,"doc_module_name":9,"category_name":35,"show_sort_weight":36,"slug":37},7,"Healthcare",40,"healthcare",{"id":39,"doc_module":4,"doc_module_name":9,"category_name":40,"show_sort_weight":41,"slug":42},8,"Research & Report",30,"research-report",{"id":44,"doc_module":4,"doc_module_name":9,"category_name":45,"show_sort_weight":46,"slug":47},9,"Religion & Spirituality",20,"religion-spirituality",{"id":46,"doc_module":4,"doc_module_name":9,"category_name":49,"show_sort_weight":46,"slug":50},"World Cup","world-cup",{"id":52,"doc_module":4,"doc_module_name":9,"category_name":53,"show_sort_weight":52,"slug":54},10,"Lifestyle","lifestyle",{"id":56,"doc_module":4,"doc_module_name":9,"category_name":57,"show_sort_weight":24,"slug":58},19,"General","general",{"code":4,"msg":60,"data":61},"ok",{"site_id":62,"language":63,"slug":64,"title":65,"keywords":66,"description":67,"schema_data":68,"social_meta":123,"head_meta":125,"extra_data":127,"updated_unix":129},105,"en","additional-mathematics-form-4-chapter-5-progressions","Additional Mathematics Form 4 Chapter 5 - Progressions","","Arithmetic progression练习资料涵盖“判定等差数列、求指定项、求首项与公差、求项数、利用三连续项求参数、以及解与项为正/负相关的应用题”。内容以多组例题与配套题目呈现，要求计算公差与通项公式Tn=a+(n−1)d，并运用等差中项性质Tn-1与Tn+1的平均关系。题目同时包含英文与马来文题干，用于训练推理与代数运算。",{"@graph":69,"@context":122},[70,84,105],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":20,"@type":76,"position":81},"https://docshare.wps.com/document/exam/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/additional-mathematics-form-4-chapter-5-progressions/138393/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/additional-mathematics-form-4-chapter-5-progressions/138393.png","ImageObject",300,407,{"name":92,"@type":93},"Sage","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-09-20","2026-08-23",true,{"@type":102,"interactionType":103,"userInteractionCount":34},"InteractionCounter",{"@type":104},"ViewAction",{"@type":106,"mainEntity":107},"FAQPage",[108,114,118],{"name":109,"@type":110,"acceptedAnswer":111},"如何判断给出的数列是否为等差数列？","Question",{"text":112,"@type":113},"计算相邻项差值并比较差值是否恒定；若所有差值相等即可判定为等差数列，并给出理由。","Answer",{"name":115,"@type":110,"acceptedAnswer":116},"已知等差数列的首项a和公差d，如何求第n项Tn？",{"text":117,"@type":113},"使用通项公式Tn = a + (n − 1)d，把n代入即可求出指定项。",{"name":119,"@type":110,"acceptedAnswer":120},"文中提到等差数列的等差中项性质如何用于求p？",{"text":121,"@type":113},"三连续项中，中间项等于两端项的平均数；用该关系把含p的表达式列式求解。","https://schema.org",{"og:url":83,"og:type":124,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":126,"canonical":83},"index,follow",{"doc_id":128,"site_id":62},138393,1787480419,{"code":4,"msg":5,"data":131},{"doc_id":128,"user_id":132,"nickname":92,"user_avatar":133,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":34,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":46,"language":139,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":140,"faqs":141,"seo_title":142,"seo_description":67,"update_tm":129,"read_time":31},687197207057,"https://ap-avatar.wpscdn.com/davatar_29158cc5080c5b710cf443261637dec0","Progressions Janjang  \n5  \n5.1 AJarnithjanmgetAictPmrogetikressions T xtbg 128o– k3   \n1. Determine whether each of the following sequences is an arithmetic progression. Give your justification. PL 2 Tentukan sama ada setiap jujukan yang berikut ialahjanjang aritmetik atau bukan. Berikan justifikasi anda.  \n| \u003Cbr>Example\u003Cbr>7, 3,–1,–5,… | (a) 12~~ ~~, ~~ ~~13~~ ~~, 14~~ ~~, ~~ ~~15~~ ~~,… |\n| --- | --- |\n| d1 = –5 –(–1) = –4 | d1 = ~~ ~~13~~ ~~ – ~~ ~~12~~ ~~ = – ~~ ~~16 |\n| d2 = –1 – 3 = –4\u003Cbr>d3 = 3 – 7 = –4 | d2 = ~~ ~~14~~ ~~ – ~~ ~~13~~ ~~ = – ~~ ~~112 |\n| An arithmetic progression because d1 = d2 = d3 = –4. | d3 = ~~ ~~15~~ ~~ – ~~ ~~14~~ ~~ = – 210 |\n|  | Not an arithmetic progression because |\n|  | d1 ≠ d2 ≠ d3. |\n| (b) –2, 3, 8, 13,… | (c) 2x + y, x,–y,–x – 2y,… |\n| d1 = 3 –(–2) = 5 | d1 = (–x – 2y)–(–y) = –x – y |\n| d2 = 8 – 3 = 5 | d2 = –y – x = –x – y |\n| d3 = 13 – 8 = 5 | d3 = x –(2x + y) = –x – y |\n| An arithmetic progression because | An arithmetic progression because |\n| d1 = d2 = d3 = 5. | d1 = d2 = d3 = –x – y. |\n\n65 © Penerbitan Pelangi Sdn. Bhd.  \n Additional Mathematics Form 4 Chapter 5 Progressions  \n2. For each of the following arithmetic progressions, determine the term stated in bracket. PL 3 Bagi setiapjanjang aritmetik berikut, tentukan sebutan yang dinyatakan dalam kurungan.  \n| \u003Cbr>Example\u003Cbr>ln x, ln 3x, ln 9x, ln 27x,… [5th term/ sebutan ke-5] a = ln x\u003Cbr>d = ln 3x – ln x = ln ~~ ~~ = ln 3\u003Cbr>T5 = ln x + 4 ln 3  Tn = a + (n – 1)d = ln 34x = ln 81x | (a) 3, 7, 11, 15,… [8th term/ sebutan ke-8]\u003Cbr>a = 3\u003Cbr>d = 7 – 3 = 4\u003Cbr>T8 = 3 + 7(4)\u003Cbr>= 31 |\n| --- | --- |\n| (b) –10,–13,–16,–19,… [16th term/ sebutan ke-16]\u003Cbr>a = –10\u003Cbr>d = –13 –(–10) = –3\u003Cbr>T16 = –10 + 15(–3)\u003Cbr>= –55 | (c) 2p, ~~ ~~3p2~~ ~~, p, ~~ ~~p2~~ ~~,… [21th term/ sebutan ke-21]\u003Cbr>a = 2p\u003Cbr>d = ~~ ~~3p2~~ ~~ – 2p = – ~~ ~~p2\u003Cbr>T21 = 2p + 201– ~~ ~~p2~~ ~~2 = 2p – 10p = –8p |\n\n3. Determine the first term and the common difference of each of the following.  PL 4 Tentukan sebutan pertama dan beza sepunya bagi setiap yang berikut.  \nExample  \nIn an arithmetic progression, the 4th and 8th terms are 16 and 56 respectively. Dalam suatu janjang aritmetik, sebutan ke-4 dan ke-8 masing-masing ialah 16 dan 56.  \nTn = a + (n – 1)d T4 = a + (4 – 1)d = 16  \na + 3d = 16 ………… 1  \nT8 = a + (8 – 1)d = 56  \na + 7d = 56 ………… 2  \n2 – 1, 4d = 40  \nd = 10  \nSubstitute d = 10 into 1, a + 3(10) = 16  \na = –14  \n\\ The first term is –14 and the common difference is 10.  \n\n| (a) In an arithmetic progression, the 3rd and 8th terms are –5 and 15 respectively.\u003Cbr>Dalam suatu janjang aritmetik, sebutan ke-3 dan ke-8 masing-masing ialah –5 dan 15.\u003Cbr>T3 = a + (3 – 1)d = –5\u003Cbr>a + 2d = –5 ……… 1\u003Cbr>T8 = a + (8 – 1)d = 15\u003Cbr>a + 7d = 15 ……… 2\u003Cbr>2 – 1, 5d = 20\u003Cbr>d = 4\u003Cbr>Substitute d = 4 into 1,\u003Cbr>a + 2(4) = –5\u003Cbr>a = –13\u003Cbr>\\ The first term is –13 and the common difference is 4. | (b) In an arithmetic progression, the 2nd and 5th terms are x and 2y respectively.\u003Cbr>Dalam suatu janjang aritmetik, sebutan ke-2 dan ke-5 masing-masing ialah x dan 2y.\u003Cbr>T2 = a + d = x………… 1\u003Cbr>T5 = a + 4d = 2y ………… 2\u003Cbr>2 – 1, 3d = 2y – xd = ~~ ~~2y3–~~ ~~x\u003Cbr>Substitute d = ~~ ~~2y3–~~ ~~x into 1,\u003Cbr>2y – x\u003Cbr>a + 3 = x\u003Cbr>2y – x\u003Cbr>a = x –\u003Cbr>3 4x – 2y\u003Cbr>=\u003Cbr>3\u003Cbr>\\ The first term is ~~ ~~4x~~ ~~–3~~ ~~2y~~ ~~ and the common\u003Cbr>difference is  2y – x 3 . |\n| --- | --- |\n\n© Penerbitan Pelangi Sdn. Bhd. 66  \nAdditional Mathematics Form 4 Chapter 5 Progressions    \n4. Find the number of terms for each of the following arithmetic progressions.  PL 4  \nCari bilangan sebutan bagi setiapjanjang aritmetik berikut.  \nExample  \n43, 31, 19,…,–29  \na = 43,  \nd = 31 – 43 = –12, Tn = –29  \nTn = a + (n – 1)d  \n–29 = 43 + (n – 1)(–12) n = ~~ ~~–2–91–243~~ ~~ + 1  \n= 7  \n\n| (a) –7, 3, 13,…, 93\u003Cbr>a = –7,\u003Cbr>d = 3 –(–7) = 10, Tn = 93\u003Cbr>93 = –7 + (n – 1)(10)\u003Cbr>n = ~~ ~~931+0~~ ~~7~~ ~~ + 1 = 11 | (b) 4p, 11p, 18p,…, 67p\u003Cbr>a = 4p,\u003Cbr>d = 11p – 4p = 7p, Tn = 67p\u003Cbr>67p = 4p + (n – 1)(7p) n =  67p – 4p","cbCaiunJ2QaVfL1X","https://ap.wps.com/l/cbCaiunJ2QaVfL1X","pdf",1435354,"English","# 5.1 判断等差数列\n# 判定并说明理由\n# 计算括号内指定项\n## 求8th、16th、21th等项\n# 求首项与公差\n# 求等差数列项数\n# 利用三连续项求p\n# 解等差数列应用题：求最小n并确定项","[{\"question\":\"如何判断给出的数列是否为等差数列？\",\"answer\":\"计算相邻项差值并比较差值是否恒定；若所有差值相等即可判定为等差数列，并给出理由。\"},{\"question\":\"已知等差数列的首项a和公差d，如何求第n项Tn？\",\"answer\":\"使用通项公式Tn = a + (n − 1)d，把n代入即可求出指定项。\"},{\"question\":\"文中提到等差数列的等差中项性质如何用于求p？\",\"answer\":\"三连续项中，中间项等于两端项的平均数；用该关系把含p的表达式列式求解。\"}]","Additional Mathematics Form 4 Chapter 5 - Progressions | PDF"]