{"id":1327,"date":"2025-12-23T13:24:48","date_gmt":"2025-12-23T12:24:48","guid":{"rendered":"http:\/\/mymatma.pl\/?p=1327"},"modified":"2025-12-23T13:24:51","modified_gmt":"2025-12-23T12:24:51","slug":"zadanie-133","status":"publish","type":"post","link":"https:\/\/mymatma.pl\/?p=1327","title":{"rendered":"Zadanie 133"},"content":{"rendered":"\n<p><strong>zadanie 17 \u2013 maj 2024 ( zadanie otwarte) (2pkt)<\/strong><\/p>\n\n\n\n<p>Ci\u0105g arytmetyczny (a<sub>n<\/sub>) jest okre\u015blony dla ka\u017cdej liczby naturalnej n\u22651 . Trzeci wyraz tego ci\u0105gu jest r\u00f3wny (\u22121) , a suma pi\u0119tnastu pocz\u0105tkowych kolejnych wyraz\u00f3w tego ci\u0105gu jest r\u00f3wna (\u2212165) . Oblicz r\u00f3\u017cnic\u0119 tego ci\u0105gu. Zapisz obliczenia.<\/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_69e22b9ed9712\">\n        <button class=\"premium-btn-odp\"\n                data-id=\"odp_69e22b9ed9712\"\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 d = \u22122.\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_69e22b9ed971e\">\n        <button class=\"premium-btn-wyjasnienie\"\n                data-id=\"wyjasnienie_69e22b9ed971e\"\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 <em>(a<sub>n<\/sub>)<\/em>, dla kt\u00f3rego:<br>\na<sub>3<\/sub> = \u22121 oraz S<sub>15<\/sub> = \u2212165.\n<\/p>\n\n<p>\nNiech a<sub>1<\/sub> = a (pierwszy wyraz ci\u0105gu), a r\u00f3\u017cnica ci\u0105gu = d.\n<\/p>\n\n<p>\nKorzystamy ze wzoru na n-ty wyraz ci\u0105gu arytmetycznego:<br>\na<sub>n<\/sub> = a + (n \u2212 1)d\n<\/p>\n\n<p>\nDla n = 3 mamy:<br>\na<sub>3<\/sub> = a + 2d = \u22121 &nbsp;&nbsp;&nbsp; (1)\n<\/p>\n\n<p>\nKorzystamy te\u017c ze wzoru na sum\u0119 n pocz\u0105tkowych wyraz\u00f3w ci\u0105gu arytmetycznego:<br>\nS<sub>n<\/sub> = (n\/2) \u00b7 (2a + (n \u2212 1)d)\n<\/p>\n\n<p>\nDla n = 15 mamy:<br>\nS<sub>15<\/sub> = (15\/2) \u00b7 (2a + 14d) = \u2212165\n<\/p>\n\n<p>\nPrzekszta\u0142camy to r\u00f3wnanie:<br>\n(15\/2) \u00b7 (2a + 14d) = \u2212165<br>\n2a + 14d = \u2212165 \u00b7 2 \/ 15<br>\n2a + 14d = \u221222 &nbsp;&nbsp;&nbsp; (2)\n<\/p>\n\n<p>\nZ r\u00f3wnania (1) wyznaczamy a:<br>\na + 2d = \u22121<br>\na = \u22121 \u2212 2d\n<\/p>\n\n<p>\nPodstawiamy a do r\u00f3wnania (2):<br>\n2(\u22121 \u2212 2d) + 14d = \u221222<br>\n\u22122 \u2212 4d + 14d = \u221222<br>\n\u22122 + 10d = \u221222<br>\n10d = \u221220<br>\nd = \u22122\n<\/p>\n\n<p>\n<strong>Wniosek:<\/strong> r\u00f3\u017cnica tego ci\u0105gu arytmetycznego jest r\u00f3wna d = \u22122.\n<\/p>\n\n        <\/div>\n    <\/div>\n    \n<\/div>\n<\/div>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>zadanie 17 \u2013 maj 2024 ( zadanie otwarte) (2pkt) Ci\u0105g arytmetyczny (an) jest okre\u015blony dla ka\u017cdej liczby naturalnej n\u22651 . Trzeci wyraz tego ci\u0105gu jest r\u00f3wny (\u22121) , a suma pi\u0119tnastu pocz\u0105tkowych kolejnych wyraz\u00f3w tego ci\u0105gu jest r\u00f3wna (\u2212165) . Oblicz r\u00f3\u017cnic\u0119 tego ci\u0105gu. Zapisz obliczenia.<\/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":[1,25,107,204,81,83],"tags":[23,106,205],"class_list":["post-1327","post","type-post","status-publish","format-standard","hentry","category-bez-kategorii","category-ciag-arytmetyczny","category-maj-2024","category-rozne","category-zadania-maturalne","category-zadania-tematami","tag-ciag-arytmetyczny","tag-maj-2024","tag-rozne"],"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 17 \u2013 maj 2024 ( zadanie otwarte) (2pkt) Ci\u0105g arytmetyczny (an) jest okre\u015blony dla ka\u017cdej liczby naturalnej n\u22651 . Trzeci wyraz tego ci\u0105gu jest r\u00f3wny (\u22121) , a suma pi\u0119tnastu pocz\u0105tkowych kolejnych wyraz\u00f3w tego ci\u0105gu jest r\u00f3wna (\u2212165) . Oblicz r\u00f3\u017cnic\u0119 tego ci\u0105gu. Zapisz obliczenia.","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1327","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=1327"}],"version-history":[{"count":1,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1327\/revisions"}],"predecessor-version":[{"id":1328,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1327\/revisions\/1328"}],"wp:attachment":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1327"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1327"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1327"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}