{"id":1164,"date":"2025-12-18T13:15:16","date_gmt":"2025-12-18T12:15:16","guid":{"rendered":"http:\/\/mymatma.pl\/?p=1164"},"modified":"2025-12-18T13:29:12","modified_gmt":"2025-12-18T12:29:12","slug":"zadanie-87","status":"publish","type":"post","link":"https:\/\/mymatma.pl\/?p=1164","title":{"rendered":"Zadanie 87"},"content":{"rendered":"\n<p><strong>zadanie 31 \u2013 sierpie\u0144 2017 (zadanie otwarte) (2pkt)<\/strong><\/p>\n\n\n\n<p>Dany jest ci\u0105g arytmetyczny (a<sub>n<\/sub>) , okre\u015blony dla n\u22651 , w kt\u00f3rym spe\u0142niona jest r\u00f3wno\u015b\u0107 a<sub>21<\/sub>+a<sub>24<\/sub>+a<sub>27<\/sub>+a<sub>30<\/sub>=100 . Oblicz sum\u0119 a<sub>25<\/sub>+a<sub>26<\/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_69e22dd9a734e\">\n        <button class=\"premium-btn-odp\"\n                data-id=\"odp_69e22dd9a734e\"\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\u0142owy wynik to 50.\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_69e22dd9a735a\">\n        <button class=\"premium-btn-wyjasnienie\"\n                data-id=\"wyjasnienie_69e22dd9a735a\"\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 warunek:<br>\na\u208d21\u208e + a\u208d24\u208e + a\u208d27\u208e + a\u208d30\u208e = 100.\n<\/p>\n\n<p>\nZauwa\u017camy, \u017ce wyrazy te tworz\u0105 pary symetryczne wzgl\u0119dem \u015brodka mi\u0119dzy 24 a 27:\n<\/p>\n\n<p>\na\u2082\u2081 i a\u2083\u2080 s\u0105 r\u00f3wno oddalone od \u015brodka \u2192 ich \u015brednia to a\u2082\u2085<br>\na\u2082\u2084 i a\u2082\u2087 s\u0105 r\u00f3wno oddalone od \u015brodka \u2192 ich \u015brednia to a\u2082\u2085\n<\/p>\n\n<p>\nZapisujemy sum\u0119 w postaci par:\n<\/p>\n\n<p>\n(a\u2082\u2081 + a\u2083\u2080) + (a\u2082\u2084 + a\u2082\u2087) = 100\n<\/p>\n\n<p>\nKa\u017cda para ma sum\u0119 r\u00f3wn\u0105 2a\u2082\u2085, wi\u0119c:\n<\/p>\n\n<p>\n2a\u2082\u2085 + 2a\u2082\u2085 = 100<br>\n4a\u2082\u2085 = 100<br>\na\u2082\u2085 = 25\n<\/p>\n\n<p>\nTeraz obliczamy a\u2082\u2086:\n<\/p>\n\n<p>\na\u2082\u2086 = a\u2082\u2085 + r\n<\/p>\n\n<p>\nR\u00f3\u017cnica r to po\u0142owa r\u00f3\u017cnicy mi\u0119dzy kolejnymi wyrazami z sumy, np.:\n<\/p>\n\n<p>\na\u2082\u2084 = a\u2082\u2085 \u2212 r<br>\na\u2082\u2087 = a\u2082\u2086 + r = a\u2082\u2085 + 2r\n<\/p>\n\n<p>\nZ pary a\u2082\u2084 + a\u2082\u2087 = 2a\u2082\u2085 = 50<br>\nTo r\u00f3wnanie nie pozwala wyznaczy\u0107 r \u2014 ale nie jest to potrzebne.\n<\/p>\n\n<p>\nSzukamy sumy:\n<\/p>\n\n<p>\na\u2082\u2085 + a\u2082\u2086 = a\u2082\u2085 + (a\u2082\u2085 + r) = 2a\u2082\u2085 + r\n<\/p>\n\n<p>\nWykorzystujemy par\u0119 a\u2082\u2084 + a\u2082\u2087:\n<\/p>\n\n<p>\na\u2082\u2084 + a\u2082\u2087 = (a\u2082\u2085 \u2212 r) + (a\u2082\u2085 + 2r) = 2a\u2082\u2085 + r = 50\n<\/p>\n\n<p>\nZatem:\n<\/p>\n\n<p>\na\u2082\u2085 + a\u2082\u2086 = 50\n<\/p>\n\n<p><strong>Odpowied\u017a: 50<\/strong><\/p>\n\n        <\/div>\n    <\/div>\n    \n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>zadanie 31 \u2013 sierpie\u0144 2017 (zadanie otwarte) (2pkt) Dany jest ci\u0105g arytmetyczny (an) , okre\u015blony dla n\u22651 , w kt\u00f3rym spe\u0142niona jest r\u00f3wno\u015b\u0107 a21+a24+a27+a30=100 . Oblicz sum\u0119 a25+a26 . 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":[25,26,139,81,83],"tags":[23,24,137],"class_list":["post-1164","post","type-post","status-publish","format-standard","hentry","category-ciag-arytmetyczny","category-obliczanie-wyrazow-ciagu","category-sierpien-2017-zadania-maturalne","category-zadania-maturalne","category-zadania-tematami","tag-ciag-arytmetyczny","tag-obliczanie-wyrazow-ciagu","tag-sierpien-2017"],"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 sierpie\u0144 2017 (zadanie otwarte) (2pkt) Dany jest ci\u0105g arytmetyczny (an) , okre\u015blony dla n\u22651 , w kt\u00f3rym spe\u0142niona jest r\u00f3wno\u015b\u0107 a21+a24+a27+a30=100 . Oblicz sum\u0119 a25+a26 . 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\/1164","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=1164"}],"version-history":[{"count":2,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1164\/revisions"}],"predecessor-version":[{"id":1170,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1164\/revisions\/1170"}],"wp:attachment":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1164"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1164"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1164"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}