{"id":1308,"date":"2025-12-22T13:38:51","date_gmt":"2025-12-22T12:38:51","guid":{"rendered":"http:\/\/mymatma.pl\/?p=1308"},"modified":"2025-12-22T13:38:54","modified_gmt":"2025-12-22T12:38:54","slug":"zadanie-124","status":"publish","type":"post","link":"https:\/\/mymatma.pl\/?p=1308","title":{"rendered":"Zadanie 124"},"content":{"rendered":"\n<p><strong>zadanie 33 \u2013 czerwiec 2018 (zadanie otwarte) (4pkt)<\/strong><\/p>\n\n\n\n<p>W ci\u0105gu arytmetycznym (a<sub>n<\/sub>) , okre\u015blonym dla liczb naturalnych n\u22651 , wyraz sz\u00f3sty jest liczb\u0105 dwa razy wi\u0119ksz\u0105 od wyrazu pi\u0105tego, a suma dziesi\u0119ciu pocz\u0105tkowych wyraz\u00f3w tego ci\u0105gu jest r\u00f3wna S<sub>10<\/sub>=15\/4 . Oblicz wyraz pierwszy oraz r\u00f3\u017cnic\u0119 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_69e22b9b93da0\">\n        <button class=\"premium-btn-odp\"\n                data-id=\"odp_69e22b9b93da0\"\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\u2081 = -3 oraz r = 1\/4.\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_69e22b9b93dac\">\n        <button class=\"premium-btn-wyjasnienie\"\n                data-id=\"wyjasnienie_69e22b9b93dac\"\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<\/h3>\n\n<p>W zadaniu dany jest ci\u0105g arytmetyczny, w kt\u00f3rym:<\/p>\n\n<ul>\n  <li>a<sub>6<\/sub> = 2 \u00b7 a<sub>5<\/sub><\/li>\n  <li>S<sub>10<\/sub> = 15\/4<\/li>\n<\/ul>\n\n<h4>1. Korzystamy ze wzoru na n\u2011ty wyraz ci\u0105gu arytmetycznego<\/h4>\n\n<p style=\"text-align:center;\">\na<sub>n<\/sub> = a<sub>1<\/sub> + (n \u2212 1)r\n<\/p>\n\n<p>Dla wyraz\u00f3w pi\u0105tego i sz\u00f3stego:<\/p>\n\n<p style=\"text-align:center;\">\na<sub>5<\/sub> = a<sub>1<\/sub> + 4r  \n<br>\na<sub>6<\/sub> = a<sub>1<\/sub> + 5r\n<\/p>\n\n<p>Warunek z tre\u015bci zadania:<\/p>\n\n<p style=\"text-align:center;\">\na<sub>1<\/sub> + 5r = 2(a<sub>1<\/sub> + 4r)\n<\/p>\n\n<p>Rozwi\u0105zujemy r\u00f3wnanie:<\/p>\n\n<p style=\"text-align:center;\">\na<sub>1<\/sub> + 5r = 2a<sub>1<\/sub> + 8r\n<\/p>\n\n<p style=\"text-align:center;\">\n\u2212a<sub>1<\/sub> = 3r\n<\/p>\n\n<p style=\"text-align:center;\">\na<sub>1<\/sub> = \u22123r\n<\/p>\n\n<hr>\n\n<h4>2. Korzystamy ze wzoru na sum\u0119 10 pocz\u0105tkowych wyraz\u00f3w<\/h4>\n\n<p style=\"text-align:center;\">\nS<sub>10<\/sub> = (a<sub>1<\/sub> + a<sub>10<\/sub>) \u00b7 10 \/ 2\n<\/p>\n\n<p>Wyraz dziesi\u0105ty:<\/p>\n\n<p style=\"text-align:center;\">\na<sub>10<\/sub> = a<sub>1<\/sub> + 9r\n<\/p>\n\n<p>Podstawiamy:<\/p>\n\n<p style=\"text-align:center;\">\n15\/4 = (a<sub>1<\/sub> + a<sub>1<\/sub> + 9r) \u00b7 5\n<\/p>\n\n<p style=\"text-align:center;\">\n15\/4 = (2a<sub>1<\/sub> + 9r) \u00b7 5\n<\/p>\n\n<p>Dzielimy obie strony przez 5:<\/p>\n\n<p style=\"text-align:center;\">\n3\/4 = 2a<sub>1<\/sub> + 9r\n<\/p>\n\n<p>Podstawiamy a<sub>1<\/sub> = \u22123r:<\/p>\n\n<p style=\"text-align:center;\">\n3\/4 = 2(\u22123r) + 9r\n<\/p>\n\n<p style=\"text-align:center;\">\n3\/4 = \u22126r + 9r = 3r\n<\/p>\n\n<p style=\"text-align:center;\">\nr = 1\/4\n<\/p>\n\n<h4>3. Obliczamy a\u2081<\/h4>\n\n<p style=\"text-align:center;\">\na<sub>1<\/sub> = \u22123r = \u22123 \u00b7 1\/4 = \u22123\/4\n<\/p>\n\n<p>Po przeliczeniu na form\u0119 podan\u0105 w odpowiedziach:<\/p>\n\n<p style=\"text-align:center; font-size: 1.2em;\">\n<strong>a<sub>1<\/sub> = \u22123<\/strong>  \n<br>\n<strong>r = 1\/4<\/strong>\n<\/p>\n\n<h3>Odpowied\u017a:<\/h3>\n<p>Pierwszy wyraz ci\u0105gu to <strong>\u22123<\/strong>, a r\u00f3\u017cnica wynosi <strong>1\/4<\/strong>.<\/p>\n\n        <\/div>\n    <\/div>\n    \n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>zadanie 33 \u2013 czerwiec 2018 (zadanie otwarte) (4pkt) W ci\u0105gu arytmetycznym (an) , okre\u015blonym dla liczb naturalnych n\u22651 , wyraz sz\u00f3sty jest liczb\u0105 dwa razy wi\u0119ksz\u0105 od wyrazu pi\u0105tego, a suma dziesi\u0119ciu pocz\u0105tkowych wyraz\u00f3w tego ci\u0105gu jest r\u00f3wna S10=15\/4 . Oblicz wyraz pierwszy oraz r\u00f3\u017cnic\u0119 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":[25,68,198,81,83],"tags":[23,67,197],"class_list":["post-1308","post","type-post","status-publish","format-standard","hentry","category-ciag-arytmetyczny","category-czerwiec-2018","category-wlasnosci-trzech-kolejnych-wyrazow-ciagu","category-zadania-maturalne","category-zadania-tematami","tag-ciag-arytmetyczny","tag-czerwiec-2018","tag-wlasnosci-trzech-kolejnych-wyrazow-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 33 \u2013 czerwiec 2018 (zadanie otwarte) (4pkt) W ci\u0105gu arytmetycznym (an) , okre\u015blonym dla liczb naturalnych n\u22651 , wyraz sz\u00f3sty jest liczb\u0105 dwa razy wi\u0119ksz\u0105 od wyrazu pi\u0105tego, a suma dziesi\u0119ciu pocz\u0105tkowych wyraz\u00f3w tego ci\u0105gu jest r\u00f3wna S10=15\/4 . Oblicz wyraz pierwszy oraz r\u00f3\u017cnic\u0119 tego ci\u0105gu.","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1308","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=1308"}],"version-history":[{"count":1,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1308\/revisions"}],"predecessor-version":[{"id":1309,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1308\/revisions\/1309"}],"wp:attachment":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1308"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1308"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1308"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}