{"id":1166,"date":"2025-12-18T13:19:46","date_gmt":"2025-12-18T12:19:46","guid":{"rendered":"http:\/\/mymatma.pl\/?p=1166"},"modified":"2025-12-18T13:19:50","modified_gmt":"2025-12-18T12:19:50","slug":"zadanie-88","status":"publish","type":"post","link":"https:\/\/mymatma.pl\/?p=1166","title":{"rendered":"Zadanie 88"},"content":{"rendered":"\n<p><strong>zadanie 31 \u2013 czerwiec 2016 (zadanie otwarte) (5pkt)<\/strong><\/p>\n\n\n\n<p>Dany jest ci\u0105g arytmetyczny (a<sub>n<\/sub>) okre\u015blony dla ka\u017cdej liczby naturalnej n\u22651 , w kt\u00f3rym a<sub>1<\/sub>+a<sub>2<\/sub>+a<sub>3<\/sub>+a<sub>4<\/sub>=2016 oraz a<sub>5<\/sub>+a<sub>6<\/sub>+a<sub>7<\/sub>+&#8230;+a<sub>12<\/sub>=2016 . Oblicz pierwszy wyraz, r\u00f3\u017cnic\u0119 oraz najmniejszy dodatni 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_69e22db1747c0\">\n        <button class=\"premium-btn-odp\"\n                data-id=\"odp_69e22db1747c0\"\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\u0142owe wyniki to: <p>a\u2081 = 126<\/p>\n<p>r = 14<\/p>\n<p>najmniejszy dodatni wyraz = 126<\/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_69e22db1747cb\">\n        <button class=\"premium-btn-wyjasnienie\"\n                data-id=\"wyjasnienie_69e22db1747cb\"\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<p>\nDany jest ci\u0105g arytmetyczny spe\u0142niaj\u0105cy dwa warunki:<br>\na\u2081 + a\u2082 + a\u2083 + a\u2084 = 2016<br>\na\u2085 + a\u2086 + &#8230; + a\u2081\u2082 = 2016\n<\/p>\n\n<p>\nNajpierw zapisujemy sum\u0119 pierwszych czterech wyraz\u00f3w:<br>\na\u2081 + (a\u2081 + r) + (a\u2081 + 2r) + (a\u2081 + 3r) = 2016<br>\n4a\u2081 + 6r = 2016 &nbsp; (1)\n<\/p>\n\n<p>\nNast\u0119pnie zapisujemy sum\u0119 wyraz\u00f3w od a\u2085 do a\u2081\u2082:<br>\na\u2085 = a\u2081 + 4r<br>\na\u2086 = a\u2081 + 5r<br>\n&#8230;<br>\na\u2081\u2082 = a\u2081 + 11r\n<\/p>\n\n<p>\nSuma o\u015bmiu kolejnych wyraz\u00f3w to:<br>\n8a\u2081 + (4 + 5 + &#8230; + 11)r = 2016\n<\/p>\n\n<p>\nSuma liczb od 4 do 11 wynosi:<br>\n4 + 11 = 15 \u2192 \u015brednia<br>\n8 liczb \u2192 8 \u00d7 15 = 120<br>\nZatem:<br>\n8a\u2081 + 120r = 2016 &nbsp; (2)\n<\/p>\n\n<p>\nOdejmujemy r\u00f3wnanie (1) pomno\u017cone przez 2 od r\u00f3wnania (2):<br>\n(8a\u2081 + 120r) \u2212 (8a\u2081 + 12r) = 2016 \u2212 4032<br>\n108r = \u22122016<br>\nr = 14\n<\/p>\n\n<p>\nPodstawiamy r = 14 do r\u00f3wnania (1):<br>\n4a\u2081 + 6\u00b714 = 2016<br>\n4a\u2081 + 84 = 2016<br>\n4a\u2081 = 1932<br>\na\u2081 = 126\n<\/p>\n\n<p>\nNajmniejszy dodatni wyraz ci\u0105gu to pierwszy wyraz, poniewa\u017c ci\u0105g jest rosn\u0105cy:<br>\n<strong>126<\/strong>\n<\/p>\n\n<p><strong>Odpowiedzi:<\/strong><br>\na\u2081 = 126<br>\nr = 14<br>\nnajmniejszy dodatni wyraz = 126<\/p>\n\n        <\/div>\n    <\/div>\n    \n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>zadanie 31 \u2013 czerwiec 2016 (zadanie otwarte) (5pkt) Dany jest ci\u0105g arytmetyczny (an) okre\u015blony dla ka\u017cdej liczby naturalnej n\u22651 , w kt\u00f3rym a1+a2+a3+a4=2016 oraz a5+a6+a7+&#8230;+a12=2016 . Oblicz pierwszy wyraz, r\u00f3\u017cnic\u0119 oraz najmniejszy dodatni wyraz ci\u0105gu (an) . Przyznaj sobie samodzielnie punkty, zgodnie z proponowan\u0105 punktacj\u0105.<\/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":[25,72,26,81,83],"tags":[23,71,24],"class_list":["post-1166","post","type-post","status-publish","format-standard","hentry","category-ciag-arytmetyczny","category-czerwiec-2016","category-obliczanie-wyrazow-ciagu","category-zadania-maturalne","category-zadania-tematami","tag-ciag-arytmetyczny","tag-czerwiec-2016","tag-obliczanie-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 31 \u2013 czerwiec 2016 (zadanie otwarte) (5pkt) Dany jest ci\u0105g arytmetyczny (an) okre\u015blony dla ka\u017cdej liczby naturalnej n\u22651 , w kt\u00f3rym a1+a2+a3+a4=2016 oraz a5+a6+a7+&#8230;+a12=2016 . Oblicz pierwszy wyraz, r\u00f3\u017cnic\u0119 oraz najmniejszy dodatni wyraz ci\u0105gu (an) . Przyznaj sobie samodzielnie punkty, zgodnie z proponowan\u0105 punktacj\u0105.","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1166","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=1166"}],"version-history":[{"count":1,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1166\/revisions"}],"predecessor-version":[{"id":1167,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1166\/revisions\/1167"}],"wp:attachment":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1166"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1166"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1166"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}