{"id":907,"date":"2025-12-05T13:19:37","date_gmt":"2025-12-05T12:19:37","guid":{"rendered":"https:\/\/mymatma.pl\/?p=907"},"modified":"2025-12-05T14:21:31","modified_gmt":"2025-12-05T13:21:31","slug":"zadanie-32","status":"publish","type":"post","link":"https:\/\/mymatma.pl\/?p=907","title":{"rendered":"Zadanie 32"},"content":{"rendered":"\n<p><strong>zadanie 32 \u2013 czerwiec 2015 ( zadanie otwarte) (4 pkt)<\/strong><br>Dany jest niesko\u0144czony rosn\u0105cy ci\u0105g arytmetyczny (a<sub>n<\/sub>) , dla n\u22651 taki, \u017ce a<sub>5<\/sub>=18 . Wyrazy a<sub>1<\/sub> , a<sub>3<\/sub> oraz a<sub>13<\/sub> tego ci\u0105gu s\u0105 odpowiednio pierwszym, drugim i trzecim wyrazem pewnego ci\u0105gu geometrycznego. Wyznacz wz\u00f3r na n -ty wyraz ci\u0105gu (a<sub>n<\/sub>) . Przyznaj sobie samodzielnie punkty, zgodnie z proponowan\u0105 punktacj\u0105.<\/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_69e22a9a0a198\">\n        <button class=\"premium-btn-odp\"\n                data-id=\"odp_69e22a9a0a198\"\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 a = 4n-2\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_69e22a9a0a1a4\">\n        <button class=\"premium-btn-wyjasnienie\"\n                data-id=\"wyjasnienie_69e22a9a0a1a4\"\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<!DOCTYPE html>\n<html lang=\"pl\">\n<head>\n    <meta charset=\"UTF-8\">\n    <meta name=\"viewport\" content=\"width=device-width, initial-scale=1.0\">\n<\/head>\n<body>\n    <p>Dany jest niesko\u0144czony rosn\u0105cy ci\u0105g arytmetyczny (a<sub>n<\/sub>) dla n\u22651, taki \u017ce a<sub>5<\/sub> = 18.  \n    Wyrazy a<sub>1<\/sub>, a<sub>3<\/sub> i a<sub>13<\/sub> tego ci\u0105gu tworz\u0105 ci\u0105g geometryczny.<\/p>\n\n    <div>\n        <p>Krok 1: Wz\u00f3r og\u00f3lny ci\u0105gu arytmetycznego:<\/p>\n        <p>a<sub>n<\/sub> = a<sub>1<\/sub> + (n-1)d<\/p>\n\n        <p>Krok 2: Warunek dla a<sub>5<\/sub>:<\/p>\n        <p>a<sub>5<\/sub> = a<sub>1<\/sub> + 4d = 18 \u2192 a<sub>1<\/sub> = 18 &#8211; 4d<\/p>\n\n        <p>Krok 3: Warunek ci\u0105gu geometrycznego:<\/p>\n        <p>Wyrazy a<sub>1<\/sub>, a<sub>3<\/sub>, a<sub>13<\/sub> tworz\u0105 ci\u0105g geometryczny:<\/p>\n        <p>a<sub>3<\/sub>\u00b2 = a<sub>1<\/sub> \u22c5 a<sub>13<\/sub><\/p>\n        <p>Podstawiamy a<sub>n<\/sub> = a<sub>1<\/sub> + (n-1)d:<\/p>\n        <p>(a<sub>1<\/sub> + 2d)\u00b2 = a<sub>1<\/sub> \u22c5 (a<sub>1<\/sub> + 12d)<\/p>\n\n        <p>Krok 4: Rozwi\u0105zujemy r\u00f3wnanie:<\/p>\n        <p>(a<sub>1<\/sub> + 2d)\u00b2 = a<sub>1<\/sub>\u00b2 + 4a<sub>1<\/sub>d + 4d\u00b2<\/p>\n        <p>Prawa strona: a<sub>1<\/sub>\u00b2 + 12a<sub>1<\/sub>d<\/p>\n        <p>R\u00f3wnanie: a<sub>1<\/sub>\u00b2 + 4a<sub>1<\/sub>d + 4d\u00b2 = a<sub>1<\/sub>\u00b2 + 12a<sub>1<\/sub>d \u2192 -8a<sub>1<\/sub>d + 4d\u00b2 = 0 \u2192 2d\u00b2 &#8211; 4a<sub>1<\/sub>d = 0 \u2192 d\u00b2 &#8211; 2a<sub>1<\/sub>d = 0 \u2192 d(d &#8211; 2a<sub>1<\/sub>) = 0<\/p>\n        <p>Nie bierzemy d=0 (ci\u0105g rosn\u0105cy), wi\u0119c:<\/p>\n        <p>d &#8211; 2a<sub>1<\/sub> = 0 \u2192 d = 2a<sub>1<\/sub><\/p>\n\n        <p>Krok 5: Podstawiamy do a<sub>1<\/sub> = 18 &#8211; 4d:<\/p>\n        <p>18 &#8211; 4d = a<sub>1<\/sub> \u2192 18 &#8211; 4(2a<sub>1<\/sub>) = a<sub>1<\/sub> \u2192 18 &#8211; 8a<sub>1<\/sub> = a<sub>1<\/sub> \u2192 18 = 9a<sub>1<\/sub> \u2192 a<sub>1<\/sub> = 2<\/p>\n\n        <p>Krok 6: Wyznaczamy d:<\/p>\n        <p>d = 2a<sub>1<\/sub> = 4<\/p>\n\n        <p>Krok 7: Wz\u00f3r na n-ty wyraz ci\u0105gu arytmetycznego:<\/p>\n        <p>a<sub>n<\/sub> = a<sub>1<\/sub> + (n-1)d = 2 + (n-1)\u22c54 = 4n &#8211; 2<\/p>\n\n        <p><strong>Odpowied\u017a:<\/strong> a<sub>n<\/sub> = 4n &#8211; 2<\/p>\n    <\/div>\n<\/body>\n<\/html>\n        <\/div>\n    <\/div>\n    \n<\/div>\n<\/div>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>zadanie 32 \u2013 czerwiec 2015 ( zadanie otwarte) (4 pkt)Dany jest niesko\u0144czony rosn\u0105cy ci\u0105g arytmetyczny (an) , dla n\u22651 taki, \u017ce a5=18 . Wyrazy a1 , a3 oraz a13 tego ci\u0105gu s\u0105 odpowiednio pierwszym, drugim i trzecim wyrazem pewnego ci\u0105gu geometrycznego. Wyznacz wz\u00f3r na n -ty wyraz ci\u0105gu (an) . Przyznaj sobie samodzielnie punkty, zgodnie [&hellip;]<\/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":[31,98,91,81,83],"tags":[29,97,93],"class_list":["post-907","post","type-post","status-publish","format-standard","hentry","category-ciag-geometryczny","category-czerwiec-2015","category-wyznaczanie-wzoru-ogolnego-ciagu","category-zadania-maturalne","category-zadania-tematami","tag-ciag-geometryczny","tag-czerwiec-2015","tag-wyznaczanie-wzoru-ogolnego-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 32 \u2013 czerwiec 2015 ( zadanie otwarte) (4 pkt)Dany jest niesko\u0144czony rosn\u0105cy ci\u0105g arytmetyczny (an) , dla n\u22651 taki, \u017ce a5=18 . Wyrazy a1 , a3 oraz a13 tego ci\u0105gu s\u0105 odpowiednio pierwszym, drugim i trzecim wyrazem pewnego ci\u0105gu geometrycznego. Wyznacz wz\u00f3r na n -ty wyraz ci\u0105gu (an) . Przyznaj sobie samodzielnie punkty, zgodnie&hellip;","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/907","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=907"}],"version-history":[{"count":2,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/907\/revisions"}],"predecessor-version":[{"id":928,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/907\/revisions\/928"}],"wp:attachment":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=907"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=907"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=907"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}