{"id":28274,"date":"2024-07-20T22:41:15","date_gmt":"2024-07-20T15:41:15","guid":{"rendered":"https:\/\/www.domainjava.com\/blog\/artikel\/jika-pada-suatu-barisan-aritmatika-memiliki-u5-dan-u25-berturut-turut-4-dan-14-maka-nilai-u-21-adalah\/"},"modified":"2024-07-20T22:41:15","modified_gmt":"2024-07-20T15:41:15","slug":"jika-pada-suatu-barisan-aritmatika-memiliki-u5-dan-u25-berturut-turut-4-dan-14-maka-nilai-u-21-adalah","status":"publish","type":"post","link":"https:\/\/www.domainjava.com\/id\/blog\/jika-pada-suatu-barisan-aritmatika-memiliki-u5-dan-u25-berturut-turut-4-dan-14-maka-nilai-u-21-adalah\/","title":{"rendered":"Jika pada suatu barisan aritmatika memiliki u5 dan u25 berturut-turut 4 dan 14, maka nilai u-21 adalah"},"content":{"rendered":"<p>Barisan aritmatika adalah barisan angka di mana setiap angka berikutnya didapatkan dengan menambahkan suatu bilangan tetap, yang kita sebut sebagai beda barisan, ke angka sebelumnya. Dalam barisan aritmatika, kita dapat menemukan suatu anggota tertentu jika kita tahu nilai suku awal (u-1), beda barisan (b), dan posisi suku tersebut.<\/p>\n<p>Berdasarkan judul artikel ini, kita tahu nilai suku ke-5 (u-5) dan suku ke-25 (u-25) sebuah barisan aritmatika, yaitu masing-masing 4 dan 14. Dari dua nilai suku tersebut, kita dapat mencari beda barisan (b) dengan rumus berikut:<\/p>\n<pre><code>b = (u25 - u5) \/ (25 - 5)<\/code><\/pre>\n<p>Kemudian dengan mengetahui nilai suku awal dan beda barisan, kita dapat mencari nilai suku ke-21 (u-21) dengan rumus berikut:<\/p>\n<pre><code>u21 = u5 + (21 - 5) * b<\/code><\/pre>\n<p>Mari kita coba hitung:<\/p>\n<p>Beda barisan (b) adalah <code>(14 - 4) \/ (25 - 5)<\/code> = 0.5. Dengan ini, kita bisa mencari nilai suku ke-21.<\/p>\n<p>Nilai u-21 adalah <code>4 + (21 - 5) * 0.5<\/code> = 12.<\/p>\n<p>Jadi, nilai suku ke-21 (u-21) dari barisan aritmatika yang memiliki suku ke-5 dan ke-25 masing-masing bernilai 4 dan 14 adalah 12.<\/p>\n<h2>Jadi, jawabannya apa?<\/h2>\n<p>Jadi, jika pada suatu barisan aritmatika memiliki suku ke-5 (u-5) dan suku ke-25 (u-25) berturut-turut 4 dan 14, maka nilai suku ke-21 (u-21) adalah 12.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Barisan aritmatika adalah barisan angka di mana setiap angka berikutnya didapatkan dengan menambahkan suatu bilangan tetap, yang kita sebut sebagai beda barisan, ke angka sebelumnya. Dalam barisan aritmatika, kita dapat menemukan suatu anggota tertentu jika kita tahu nilai suku awal (u-1), beda barisan (b), dan posisi suku tersebut. Berdasarkan judul artikel ini, kita tahu nilai &#8230; <a title=\"Jika pada suatu barisan aritmatika memiliki u5 dan u25 berturut-turut 4 dan 14, maka nilai u-21 adalah\" class=\"read-more\" href=\"https:\/\/www.domainjava.com\/id\/blog\/jika-pada-suatu-barisan-aritmatika-memiliki-u5-dan-u25-berturut-turut-4-dan-14-maka-nilai-u-21-adalah\/\" aria-label=\"Read more about Jika pada suatu barisan aritmatika memiliki u5 dan u25 berturut-turut 4 dan 14, maka nilai u-21 adalah\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"class_list":["post-28274","post","type-post","status-publish","format-standard","hentry","category-wawasan"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.3 (Yoast SEO v27.7) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>Jika pada suatu barisan aritmatika memiliki u5 dan u25 berturut-turut 4 dan 14, maka nilai u-21 adalah - DomainJava Blog<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.domainjava.com\/id\/blog\/jika-pada-suatu-barisan-aritmatika-memiliki-u5-dan-u25-berturut-turut-4-dan-14-maka-nilai-u-21-adalah\/\" \/>\n<meta property=\"og:locale\" content=\"id_ID\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Jika pada suatu barisan aritmatika memiliki u5 dan u25 berturut-turut 4 dan 14, maka nilai u-21 adalah\" \/>\n<meta property=\"og:description\" content=\"Barisan aritmatika adalah barisan angka di mana setiap angka berikutnya didapatkan dengan menambahkan suatu bilangan tetap, yang kita sebut sebagai beda barisan, ke angka sebelumnya. 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