{"id":876,"date":"2025-12-05T10:49:42","date_gmt":"2025-12-05T09:49:42","guid":{"rendered":"https:\/\/mymatma.pl\/?p=876"},"modified":"2025-12-05T11:15:05","modified_gmt":"2025-12-05T10:15:05","slug":"zadanie-25","status":"publish","type":"post","link":"https:\/\/mymatma.pl\/?p=876","title":{"rendered":"Zadanie 25"},"content":{"rendered":"\n<p><strong>zadanie 31 \u2013 czerwiec 2013 (zadanie otwarte) (2 pkt)<\/strong><\/p>\n\n\n\n<p>Niesko\u0144czony ci\u0105g geometryczny (a<sub>n<\/sub>) jest okre\u015blony wzorem a<sub>n<\/sub>=7\u22c53<sup>n+1<\/sup> , dla n\u22651 . Oblicz iloraz q tego ci\u0105gu. Przyznaj sobie samodzielnie punkty, zgodnie z proponowan\u0105 punktacj\u0105 ( 0 pkt, 1 pkt, 2 pkt)<\/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_69e22a75027e2\">\n        <button class=\"premium-btn-odp\"\n                data-id=\"odp_69e22a75027e2\"\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 q=3.\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_69e22a75027ee\">\n        <button class=\"premium-btn-wyjasnienie\"\n                data-id=\"wyjasnienie_69e22a75027ee\"\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<div class=\"card\" role=\"article\" style=\"font-family:system-ui,-apple-system,Segoe UI,Roboto,'Helvetica Neue',Arial;line-height:1.5;color:#111;padding:18px;background:#fff;border:1px solid #e2e6ea;border-radius:10px;max-width:760px;margin:16px auto;box-shadow:0 6px 20px rgba(18,38,63,0.06)\">\n\n  <p>Dany jest niesko\u0144czony ci\u0105g geometryczny <code>(a<sub>n<\/sub>)<\/code> okre\u015blony wzorem:<\/p>\n  <p style=\"background:#fbfdff;padding:8px;border-radius:6px;border:1px solid #eef6ff;display:inline-block\">\n    a<sub>n<\/sub> = 7 \u00b7 3<sup>n+1<\/sup>, &nbsp; n \u2265 1\n  <\/p>\n\n  <p><strong>Krok 1 \u2014 wyznacz iloraz q ci\u0105gu:<\/strong><\/p>\n  <p>Iloraz q mo\u017cna obliczy\u0107, dziel\u0105c kolejny wyraz przez poprzedni:<\/p>\n  <p style=\"background:#fbfdff;padding:8px;border-radius:6px;border:1px solid #eef6ff;display:inline-block\">\n    q = a<sub>n+1<\/sub> \/ a<sub>n<\/sub>\n  <\/p>\n\n  <p><strong>Krok 2 \u2014 podstaw wz\u00f3r:<\/strong><\/p>\n  <p style=\"background:#fbfdff;padding:8px;border-radius:6px;border:1px solid #eef6ff;display:inline-block\">\n    a<sub>n+1<\/sub> \/ a<sub>n<\/sub> = [7\u00b73<sup>(n+1)+1<\/sup>] \/ [7\u00b73<sup>n+1<\/sup>] = 3\n  <\/p>\n\n  <p><strong>Krok 3 \u2014 wniosek:<\/strong><\/p>\n  <p>Iloraz q jest sta\u0142y i dodatni, wi\u0119c:<\/p>\n\n  <div style=\"margin-top:14px;padding:12px;border-radius:8px;background:#f0faf0;border:1px solid #d8f1d8;color:#066a06;font-weight:600\">\n    Iloraz q = <strong>3<\/strong>\n  <\/div>\n\n  <hr style=\"margin:22px 0;border:none;border-top:1px solid #eee\">\n\n  <h3 style=\"font-size:1.1rem;margin:10px 0 6px 0\">Samodzielna punktacja<\/h3>\n\n  <p>Zgodnie z proponowan\u0105 punktacj\u0105 (0 pkt, 1 pkt, 2 pkt) oce\u0144 swoje rozwi\u0105zanie:<\/p>\n\n  <ul style=\"line-height:1.6;margin-left:18px\">\n    <li><strong>2 pkt<\/strong> \u2014 poprawnie obliczony iloraz q.<\/li>\n    <li><strong>1 pkt<\/strong> \u2014 prawid\u0142owo zapisano wz\u00f3r, ale pope\u0142niono drobny b\u0142\u0105d w obliczeniach.<\/li>\n    <li><strong>0 pkt<\/strong> \u2014 wynik niepoprawny lub zadanie nie rozwi\u0105zane.<\/li>\n  <\/ul>\n\n<\/div>\n\n        <\/div>\n    <\/div>\n    \n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>zadanie 31 \u2013 czerwiec 2013 (zadanie otwarte) (2 pkt) Niesko\u0144czony ci\u0105g geometryczny (an) jest okre\u015blony wzorem an=7\u22c53n+1 , dla n\u22651 . Oblicz iloraz q tego ci\u0105gu. Przyznaj sobie samodzielnie punkty, zgodnie z proponowan\u0105 punktacj\u0105 ( 0 pkt, 1 pkt, 2 pkt)<\/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":[31,78,49,81,83],"tags":[29,77,52],"class_list":["post-876","post","type-post","status-publish","format-standard","hentry","category-ciag-geometryczny","category-czerwiec-2013","category-iloraz-ciagu","category-zadania-maturalne","category-zadania-tematami","tag-ciag-geometryczny","tag-czerwiec-2013","tag-iloraz-ciagu"],"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 31 \u2013 czerwiec 2013 (zadanie otwarte) (2 pkt) Niesko\u0144czony ci\u0105g geometryczny (an) jest okre\u015blony wzorem an=7\u22c53n+1 , dla n\u22651 . Oblicz iloraz q tego ci\u0105gu. Przyznaj sobie samodzielnie punkty, zgodnie z proponowan\u0105 punktacj\u0105 ( 0 pkt, 1 pkt, 2 pkt)","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/876","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=876"}],"version-history":[{"count":1,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/876\/revisions"}],"predecessor-version":[{"id":877,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/876\/revisions\/877"}],"wp:attachment":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=876"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=876"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=876"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}