{"id":1408,"date":"2025-12-31T08:03:58","date_gmt":"2025-12-31T07:03:58","guid":{"rendered":"http:\/\/mymatma.pl\/?p=1408"},"modified":"2025-12-31T08:05:05","modified_gmt":"2025-12-31T07:05:05","slug":"zadanie-142","status":"publish","type":"post","link":"https:\/\/mymatma.pl\/?p=1408","title":{"rendered":"Zadanie 142"},"content":{"rendered":"\n<p><strong>zadanie 32 &#8211; sierpie\u0144 2019 (zadanie otwarte) (4pkt)<\/strong><\/p>\n\n\n\n<p>W ci\u0105gu arytmetycznym (a<sub>1<\/sub>,a<sub>2<\/sub>,&#8230;,a<sub>39<\/sub>,a<sub>40<\/sub>) suma wyraz\u00f3w tego ci\u0105gu o numerach parzystych jest r\u00f3wna 1340 , a suma wyraz\u00f3w ci\u0105gu o numerach nieparzystych jest r\u00f3wna 1400 . Wyznacz ostatni wyraz tego ci\u0105gu arytmetycznego.<\/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_69e22b4f97043\">\n        <button class=\"premium-btn-odp\"\n                data-id=\"odp_69e22b4f97043\"\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            \nOstatni wyraz tego ci\u0105gu arytmetycznego to <strong>a<sub>40<\/sub> = 10<\/strong>.\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_69e22b4f97051\">\n        <button class=\"premium-btn-wyjasnienie\"\n                data-id=\"wyjasnienie_69e22b4f97051\"\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<h2>Wyja\u015bnienie<\/h2>\n\n<p>W ci\u0105gu arytmetycznym mamy 40 wyraz\u00f3w: <code>a\u2081, a\u2082, ..., a\u2084\u2080<\/code>.<\/p>\n\n<h3>1. Podzia\u0142 wyraz\u00f3w na parzyste i nieparzyste<\/h3>\n<ul>\n  <li>Wyrazy nieparzyste: <code>a\u2081, a\u2083, ..., a\u2083\u2089<\/code> \u2013 jest ich 20.<\/li>\n  <li>Wyrazy parzyste: <code>a\u2082, a\u2084, ..., a\u2084\u2080<\/code> \u2013 r\u00f3wnie\u017c 20.<\/li>\n<\/ul>\n\n<h3>2. Oznaczenia<\/h3>\n<p>Niech pierwszy wyraz to <code>a\u2081<\/code>, a r\u00f3\u017cnica ci\u0105gu to <code>r<\/code>.<\/p>\n\n<h3>3. Suma wyraz\u00f3w nieparzystych<\/h3>\n<p>Wyraz 39. ma posta\u0107: <code>a\u2083\u2089 = a\u2081 + 38r<\/code>.<\/p>\n\n<p>\nSuma 20 wyraz\u00f3w nieparzystych:  \n<br>\n<code>S<sub>niep<\/sub> = (20\/2)(a\u2081 + a\u2081 + 38r) = 20a\u2081 + 380r = 1400<\/code>\n<\/p>\n\n<h3>4. Suma wyraz\u00f3w parzystych<\/h3>\n<p>Wyraz 2. to <code>a\u2082 = a\u2081 + r<\/code>, a wyraz 40. to <code>a\u2084\u2080 = a\u2081 + 39r<\/code>.<\/p>\n\n<p>\nSuma 20 wyraz\u00f3w parzystych:  \n<br>\n<code>S<sub>parz<\/sub> = (20\/2)(a\u2081 + r + a\u2081 + 39r) = 20a\u2081 + 400r = 1340<\/code>\n<\/p>\n\n<h3>5. Uk\u0142ad r\u00f3wna\u0144<\/h3>\n\n<pre>\n20a\u2081 + 380r = 1400  \n20a\u2081 + 400r = 1340\n<\/pre>\n\n<p>Odejmujemy pierwsze r\u00f3wnanie od drugiego:<\/p>\n\n<p>\n<code>20r = -60<\/code><br>\n<code>r = -3<\/code>\n<\/p>\n\n<h3>6. Wyznaczenie pierwszego wyrazu<\/h3>\n\n<p>\n<code>20a\u2081 + 380(-3) = 1400<\/code><br>\n<code>20a\u2081 - 1140 = 1400<\/code><br>\n<code>20a\u2081 = 2540<\/code><br>\n<code>a\u2081 = 127<\/code>\n<\/p>\n\n<h3>7. Obliczenie ostatniego wyrazu<\/h3>\n\n<p>\n<code>a\u2084\u2080 = a\u2081 + 39r = 127 + 39\u00b7(-3) = 127 - 117 = 10<\/code>\n<\/p>\n\n<h3>Wniosek<\/h3>\n\n<p><strong>Ostatni wyraz ci\u0105gu arytmetycznego wynosi: a\u2084\u2080 = 10.<\/strong><\/p>\n\n        <\/div>\n    <\/div>\n    \n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>zadanie 32 &#8211; sierpie\u0144 2019 (zadanie otwarte) (4pkt) W ci\u0105gu arytmetycznym (a1,a2,&#8230;,a39,a40) suma wyraz\u00f3w tego ci\u0105gu o numerach parzystych jest r\u00f3wna 1340 , a suma wyraz\u00f3w ci\u0105gu o numerach nieparzystych jest r\u00f3wna 1400 . Wyznacz ostatni wyraz tego ci\u0105gu arytmetycznego.<\/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":[25,204,62,81,83],"tags":[23,205,61],"class_list":["post-1408","post","type-post","status-publish","format-standard","hentry","category-ciag-arytmetyczny","category-rozne","category-sierpien-2019","category-zadania-maturalne","category-zadania-tematami","tag-ciag-arytmetyczny","tag-rozne","tag-sierpien-2019"],"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 &#8211; sierpie\u0144 2019 (zadanie otwarte) (4pkt) W ci\u0105gu arytmetycznym (a1,a2,&#8230;,a39,a40) suma wyraz\u00f3w tego ci\u0105gu o numerach parzystych jest r\u00f3wna 1340 , a suma wyraz\u00f3w ci\u0105gu o numerach nieparzystych jest r\u00f3wna 1400 . Wyznacz ostatni wyraz tego ci\u0105gu arytmetycznego.","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1408","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=1408"}],"version-history":[{"count":2,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1408\/revisions"}],"predecessor-version":[{"id":1418,"href":"https:\/\/mymatma.pl\/index.php?rest_route=\/wp\/v2\/posts\/1408\/revisions\/1418"}],"wp:attachment":[{"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1408"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1408"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mymatma.pl\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1408"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}