{"id":1191,"date":"2025-12-19T09:45:12","date_gmt":"2025-12-19T08:45:12","guid":{"rendered":"https:\/\/mymatma.pl\/?p=1191"},"modified":"2025-12-19T09:45:17","modified_gmt":"2025-12-19T08:45:17","slug":"zadanie-99","status":"publish","type":"post","link":"https:\/\/mymatma.pl\/?p=1191","title":{"rendered":"Zadanie 99"},"content":{"rendered":"\n<p><strong>zadanie 30 \u2013 maj 2022 (zadanie otwarte) (2pkt)<\/strong><\/p>\n\n\n\n<p>W ci\u0105gu arytmetycznym (a<sub>n<\/sub>) , okre\u015blonym dla ka\u017cdej liczby naturalnej n\u22651 , a<sub>1<\/sub>=\u22121 i a<sub>4<\/sub>=8 . Oblicz sum\u0119 stu pocz\u0105tkowych kolejnych wyraz\u00f3w tego ci\u0105gu.<\/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_69e22bbb21e50\">\n        <button class=\"premium-btn-odp\"\n                data-id=\"odp_69e22bbb21e50\"\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            \nPrawid\u0142owa odpowied\u017a to 49 450.\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_69e22bbb21e5c\">\n        <button class=\"premium-btn-wyjasnienie\"\n                data-id=\"wyjasnienie_69e22bbb21e5c\"\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<h3>Wyja\u015bnienie zadania \u2014 suma 100 wyraz\u00f3w ci\u0105gu arytmetycznego<\/h3>\n<ol>\n  <li>\n    <strong>Dane z tre\u015bci zadania:<\/strong>\n    <ul>\n      <li>a<sub>1<\/sub> = -1<\/li>\n      <li>a<sub>4<\/sub> = 8<\/li>\n    <\/ul>\n  <\/li>\n  <li>\n    <strong>Wyznacz r\u00f3\u017cnic\u0119 ci\u0105gu:<\/strong><br>\n    Wz\u00f3r og\u00f3lny: a<sub>n<\/sub> = a<sub>1<\/sub> + (n &#8211; 1)\u00b7r<br>\n    Podstawiaj\u0105c n = 4:<br>\n    a<sub>4<\/sub> = a<sub>1<\/sub> + 3r = -1 + 3r = 8<br>\n    3r = 9 \u21d2 r = 3\n  <\/li>\n  <li>\n    <strong>Oblicz setny wyraz ci\u0105gu:<\/strong><br>\n    a<sub>100<\/sub> = a<sub>1<\/sub> + 99r = -1 + 99\u00b73 = -1 + 297 = 296\n  <\/li>\n  <li>\n    <strong>Skorzystaj ze wzoru na sum\u0119 n pocz\u0105tkowych wyraz\u00f3w:<\/strong><br>\n    S<sub>n<\/sub> = (a<sub>1<\/sub> + a<sub>n<\/sub>) \u00b7 n \/ 2<br>\n    S<sub>100<\/sub> = (-1 + 296) \u00b7 100 \/ 2 = 295 \u00b7 50 = 14 750\n  <\/li>\n  <li>\n    <strong>Uwaga:<\/strong> W tym zadaniu chodzi o <em>sto pocz\u0105tkowych kolejnych wyraz\u00f3w<\/em>, czyli od a<sub>1<\/sub> do a<sub>100<\/sub>. Jednak wynik 14 750 nie zgadza si\u0119 z podanym rozwi\u0105zaniem 49 450. Sprawd\u017amy ponownie:\n    <br><br>\n    Mo\u017cna te\u017c policzy\u0107 przez wz\u00f3r: S<sub>n<\/sub> = n\/2 \u00b7 (2a<sub>1<\/sub> + (n &#8211; 1)r)<br>\n    S<sub>100<\/sub> = 100\/2 \u00b7 (2\u00b7(-1) + 99\u00b73)<br>\n    = 50 \u00b7 (-2 + 297)<br>\n    = 50 \u00b7 295 = 14 750\n  <\/li>\n<\/ol>\n<p><strong>Wniosek:<\/strong> Prawid\u0142owa suma 100 pocz\u0105tkowych wyraz\u00f3w tego ci\u0105gu wynosi <u>14 750<\/u>. Je\u015bli w tre\u015bci zadania podano wynik 49 450, to najprawdopodobniej chodzi\u0142o o inne dane lub liter\u00f3wk\u0119 w zadaniu.<\/p>\n\n        <\/div>\n    <\/div>\n    \n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>zadanie 30 \u2013 maj 2022 (zadanie otwarte) (2pkt) W ci\u0105gu arytmetycznym (an) , okre\u015blonym dla ka\u017cdej liczby naturalnej n\u22651 , a1=\u22121 i a4=8 . Oblicz sum\u0119 stu pocz\u0105tkowych kolejnych wyraz\u00f3w tego ci\u0105gu.<\/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":[25,58,160,81,83],"tags":[],"class_list":["post-1191","post","type-post","status-publish","format-standard","hentry","category-ciag-arytmetyczny","category-maj-2022","category-obliczanie-roznicy-ciagu","category-zadania-maturalne","category-zadania-tematami"],"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 30 \u2013 maj 2022 (zadanie otwarte) (2pkt) W ci\u0105gu arytmetycznym (an) , okre\u015blonym dla ka\u017cdej liczby naturalnej n\u22651 , a1=\u22121 i a4=8 . Oblicz sum\u0119 stu pocz\u0105tkowych kolejnych wyraz\u00f3w tego ci\u0105gu.","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1191","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=1191"}],"version-history":[{"count":1,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1191\/revisions"}],"predecessor-version":[{"id":1192,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1191\/revisions\/1192"}],"wp:attachment":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1191"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1191"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1191"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}