{"id":1419,"date":"2025-12-31T08:09:03","date_gmt":"2025-12-31T07:09:03","guid":{"rendered":"http:\/\/mymatma.pl\/?p=1419"},"modified":"2025-12-31T08:09:07","modified_gmt":"2025-12-31T07:09:07","slug":"zadanie-143","status":"publish","type":"post","link":"https:\/\/mymatma.pl\/?p=1419","title":{"rendered":"Zadanie 143"},"content":{"rendered":"\n<p><strong>zadanie 32 \u2013 maj 2019 (zadanie otwarte) (4pkt)<\/strong><\/p>\n\n\n\n<p>Ci\u0105g arytmetyczny jest okre\u015blony dla ka\u017cdej liczby naturalnej . R\u00f3\u017cnic\u0105 tego ci\u0105gu jest liczba r=\u22124 , a \u015brednia arytmetyczna pocz\u0105tkowych sze\u015bciu wyraz\u00f3w tego ci\u0105gu: a<sub>1<\/sub>,a<sub>2<\/sub>,a<sub>3<\/sub>,a<sub>4<\/sub>,a<sub>5<\/sub>,a<sub>6<\/sub> jest r\u00f3wna 16.<\/p>\n\n\n\n<p>a) Oblicz pierwszy wyraz tego ci\u0105gu.<\/p>\n\n\n\n<p>b) Oblicz liczb\u0119 k , dla kt\u00f3rej a<sub>k<\/sub>=\u221278 .<\/p>\n\n\n\n<div class=\"wp-block-columns is-layout-flex wp-container-core-columns-is-layout-28f84493 wp-block-columns-is-layout-flex\">\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\">\n    <div class=\"premium-wrapper-odp\" id=\"odp_69e22d90638b0\">\n        <button class=\"premium-btn-odp\"\n                data-id=\"odp_69e22d90638b0\"\n                style=\"margin-top:10px; padding:12px 20px; background:#28a745; color:#fff; border-radius:25px; border:none; cursor:pointer; font-weight:bold;\">\n            Poka\u017c odpowied\u017a\n        <\/button>\n\n        <div class=\"premium-content-odp\" style=\"display:none; padding:15px; border:1px solid #28a745; margin-top:10px; background:#e6ffe6; border-radius:25px;\">\n            \n<p>a) Pierwszy wyraz ci\u0105gu: <strong>a\u2081 = 26<\/strong><\/p> <p>b) Liczba k spe\u0142niaj\u0105ca warunek a\u2096 = \u221278: <strong>k = 27<\/strong><\/p>\n        <\/div>\n    <\/div>\n    \n<\/div>\n\n\n\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\">\n    <div class=\"premium-wrapper-wyjasnienie\" id=\"wyjasnienie_69e22d90638bd\">\n        <button class=\"premium-btn-wyjasnienie\"\n                data-id=\"wyjasnienie_69e22d90638bd\"\n                style=\"margin-top:10px; padding:12px 20px; background:#28a745; color:#fff; border-radius:25px; border:none; cursor:pointer; font-weight:bold;\">\n            Poka\u017c wyja\u015bnienie\n        <\/button>\n\n        <div class=\"premium-content-wyjasnienie\" style=\"display:none; padding:15px; border:1px solid #28a745; margin-top:10px; background:#e6ffe6; border-radius:25px;\">\n            \n<h2>Wyja\u015bnienie<\/h2>\n\n<p>W zadaniu dany jest ci\u0105g arytmetyczny o r\u00f3\u017cnicy <code>r = -4<\/code>. \n\u015arednia arytmetyczna pierwszych sze\u015bciu wyraz\u00f3w wynosi <code>16<\/code>.<\/p>\n\n<h3>Krok 1: Obliczenie pierwszego wyrazu<\/h3>\n\n<p>Pierwsze sze\u015b\u0107 wyraz\u00f3w ci\u0105gu to:<\/p>\n<p><code>a\u2081, a\u2081 - 4, a\u2081 - 8, a\u2081 - 12, a\u2081 - 16, a\u2081 - 20<\/code><\/p>\n\n<p>Suma tych wyraz\u00f3w:<\/p>\n<p><code>S\u2086 = 6a\u2081 - (4 + 8 + 12 + 16 + 20) = 6a\u2081 - 60<\/code><\/p>\n\n<p>\u015arednia arytmetyczna wynosi 16, wi\u0119c:<\/p>\n<p><code>(6a\u2081 - 60) \/ 6 = 16<\/code><\/p>\n\n<p>Po przekszta\u0142ceniu:<\/p>\n<p><code>6a\u2081 - 60 = 96<\/code><\/p>\n<p><code>6a\u2081 = 156<\/code><\/p>\n<p><strong>a\u2081 = 26<\/strong><\/p>\n\n<h3>Krok 2: Wyznaczenie k, dla kt\u00f3rego a\u2096 = -78<\/h3>\n\n<p>Korzystamy ze wzoru na n-ty wyraz ci\u0105gu arytmetycznego:<\/p>\n<p><code>a\u2096 = a\u2081 + (k - 1)r<\/code><\/p>\n\n<p>Podstawiamy dane:<\/p>\n<p><code>-78 = 26 + (k - 1)(-4)<\/code><\/p>\n\n<p>Przekszta\u0142camy:<\/p>\n<p><code>-78 - 26 = -4(k - 1)<\/code><\/p>\n<p><code>-104 = -4(k - 1)<\/code><\/p>\n<p><code>k - 1 = 26<\/code><\/p>\n<p><strong>k = 27<\/strong><\/p>\n\n<h3>Wynik ko\u0144cowy<\/h3>\n<ul>\n  <li><strong>a\u2081 = 26<\/strong><\/li>\n  <li><strong>k = 27<\/strong><\/li>\n<\/ul>\n\n        <\/div>\n    <\/div>\n    \n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>zadanie 32 \u2013 maj 2019 (zadanie otwarte) (4pkt) Ci\u0105g arytmetyczny jest okre\u015blony dla ka\u017cdej liczby naturalnej . R\u00f3\u017cnic\u0105 tego ci\u0105gu jest liczba r=\u22124 , a \u015brednia arytmetyczna pocz\u0105tkowych sze\u015bciu wyraz\u00f3w tego ci\u0105gu: a1,a2,a3,a4,a5,a6 jest r\u00f3wna 16. a) Oblicz pierwszy wyraz tego ci\u0105gu. b) Oblicz liczb\u0119 k , dla kt\u00f3rej ak=\u221278 .<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"saved_in_kubio":false,"_uag_custom_page_level_css":"","_themeisle_gutenberg_block_has_review":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[25,64,204,81,83],"tags":[23,63,205],"class_list":["post-1419","post","type-post","status-publish","format-standard","hentry","category-ciag-arytmetyczny","category-maj-2019","category-rozne","category-zadania-maturalne","category-zadania-tematami","tag-ciag-arytmetyczny","tag-maj-2019","tag-rozne"],"jetpack_featured_media_url":"","uagb_featured_image_src":{"full":false,"thumbnail":false,"medium":false,"medium_large":false,"large":false,"1536x1536":false,"2048x2048":false,"kubio-fullhd":false,"woocommerce_thumbnail":false,"woocommerce_single":false,"woocommerce_gallery_thumbnail":false},"uagb_author_info":{"display_name":"adminmymatma","author_link":"https:\/\/mymatma.pl\/?author=1"},"uagb_comment_info":0,"uagb_excerpt":"zadanie 32 \u2013 maj 2019 (zadanie otwarte) (4pkt) Ci\u0105g arytmetyczny jest okre\u015blony dla ka\u017cdej liczby naturalnej . R\u00f3\u017cnic\u0105 tego ci\u0105gu jest liczba r=\u22124 , a \u015brednia arytmetyczna pocz\u0105tkowych sze\u015bciu wyraz\u00f3w tego ci\u0105gu: a1,a2,a3,a4,a5,a6 jest r\u00f3wna 16. a) Oblicz pierwszy wyraz tego ci\u0105gu. b) Oblicz liczb\u0119 k , dla kt\u00f3rej ak=\u221278 .","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1419","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1419"}],"version-history":[{"count":1,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1419\/revisions"}],"predecessor-version":[{"id":1423,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1419\/revisions\/1423"}],"wp:attachment":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1419"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1419"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1419"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}