{"id":160243,"date":"2020-08-06T19:05:00","date_gmt":"2020-08-06T22:05:00","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=160243"},"modified":"2023-07-06T05:55:41","modified_gmt":"2023-07-06T08:55:41","slug":"problema-numeros-em-progressao-aritmetica","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-numeros-em-progressao-aritmetica\/","title":{"rendered":".Problema: N\u00fameros em progress\u00e3o aritm\u00e9tica"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\" style=\"line-height:25px;padding-bottom:15px\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 2\u00ba ano do E. M.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nSabendo que <\/p>\n<ul>\n<li>[tex](x + 2y\\;,\\; 3x \u2212 5y\\;,\\; 8x \u2212 2y\\;,\\; 11x \u2212 7y + 2z)[\/tex] <\/li>\n<\/ul>\n<p>\u00e9 uma progress\u00e3o aritm\u00e9tica com o \u00faltimo termo igual a [tex]\u2212127[\/tex], <strong>calcule [tex]x\\cdot y\\cdot z[\/tex]<\/strong>.<\/p>\n<p style=\"font-size: 12px;line-height:12px; text-align:right\">(<strong>ITA<\/strong>, 2012 &#8211; Adaptado) <\/p>\n<\/div>\n<div class=\"cb36 g1n\" style=\"margin:40px;line-height:30px; width:57%; margin-left:auto; margin-right:auto; \"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" style=\"border:none\"  alt=\"explicador_p\" width=\"70\" height=\"84\" class=\"alignright size-full wp-image-1001\" \/> <\/p>\n<p class=\"c3 ti3\" style=\"line-height:25px;margin:10px\"> Lembretes<\/p>\n<p>Dada uma progress\u00e3o aritm\u00e9tica [tex](a_{1}, a_{2}, a_{3}, \\dots)[\/tex] de raz\u00e3o [tex]r[\/tex]:<\/p>\n<ul>[tex]\\textcolor{#800000}{(a)}[\/tex] O termo geral, [tex]a_{n}[\/tex], \u00e9 dado por   [tex]\\boxed{a_{n}=a_{1}+ (n-1)\\cdot r}[\/tex].<br \/>\n\t[tex]\\textcolor{#800000}{(b)}[\/tex]  [tex]r=a_2-a_1=a_3-a_2=a_4-a_3=\\cdots[\/tex].<br \/>\n\t[tex]\\textcolor{#800000}{(c)}[\/tex]  [tex]a_k=\\dfrac{a_{k+1}+a_{k-1}}{2}[\/tex], para todo [tex]k\\geq 2[\/tex]. <\/ul>\n<\/div>\n<div class=\"g1n\" style=\"line-height:28px;\">\n<p class=\"tit\" style=\"line-height:28px;margin:10px\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 1px; margin-bottom: 20px;border: 0 none\"\/>\nDe acordo com a nota\u00e7\u00e3o dos <strong>Lembretes<\/strong>, vamos denotar a progress\u00e3o dada no problema como [tex](a_{1}, a_{2}, a_{3}, a_{4})\\,.[\/tex]<\/p>\n<ul>\n<li>Como [tex]a_4=-127[\/tex] temos<\/li>\n<p>[tex]\\qquad 11x-7y+2z=-127.\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\textcolor{#800000}{(i)}[\/tex]<\/p>\n<li>Pelo Lembrete [tex]\\textcolor{#800000}{(b)}[\/tex] temos<\/li>\n<p>[tex]\\qquad r=a_2-a_1=(3x-5y)-(x+2y)[\/tex]<br \/>\n[tex]\\qquad r=2x-7y.\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\textcolor{#800000}{(ii)}[\/tex]<\/p>\n<li>Do Lembrete [tex]\\textcolor{#800000}{(a)}[\/tex], com [tex]n=4[\/tex], obtemos<\/li>\n<p>[tex]\\qquad a_4=a_1+(4-1)\\cdot r[\/tex]<br \/>\n[tex]\\qquad -127=(x+2y)+3\\cdot (2x-7y)[\/tex]<br \/>\n[tex]\\qquad -127=x+2y+6x-21y[\/tex]<br \/>\n[tex]\\qquad 7x-19y=-127.\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\textcolor{#800000}{(iii)}[\/tex]<\/p>\n<li>Usando o Lembrete [tex]\\textcolor{#800000}{(c)}[\/tex] para [tex]k=3[\/tex], vemos que<\/li>\n<p>[tex]\\qquad a_3=\\dfrac{a_4+a_2}{2}[\/tex]<br \/>\n[tex]\\qquad (8x-2y)=\\dfrac{(11x-7y+2z)+(3x-5y)}{2}[\/tex]<br \/>\n[tex]\\qquad 2\\cdot (8x-2y)=(11x-7y+2z)+(3x-5y)[\/tex]<br \/>\n[tex]\\qquad 16x-4y=14x-12y+2z[\/tex]<br \/>\n[tex]\\qquad 2x+8y-2z=0.\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\textcolor{#800000}{(iv)}[\/tex]<\/p>\n<li>De [tex]\\textcolor{#800000}{(iii)}[\/tex], vem que<\/li>\n<p>[tex]\\qquad 7x-19y=-127[\/tex]<br \/>\n[tex]\\qquad 7x=-127+19y[\/tex]<br \/>\n[tex]\\qquad x=\\dfrac{-127+19y}{7}.\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\textcolor{#800000}{(v)}[\/tex]<\/p>\n<li>Substituindo [tex]\\textcolor{#800000}{(v)}[\/tex] em [tex]\\textcolor{#800000}{(iv)}[\/tex] obtemos<\/li>\n<p>[tex]\\qquad 2x+8y-2z=0[\/tex]<br \/>\n[tex]\\qquad 2\\cdot \\bigg(\\dfrac{-127+19y}{7}\\bigg)+8y-2z=0[\/tex]<br \/>\n[tex]\\qquad 2\\cdot(-127+19y)+56y-14z=0[\/tex]<br \/>\n[tex]\\qquad -254+38y+56y-14z=0[\/tex]<br \/>\n[tex]\\qquad 14z=-254+94y[\/tex]<br \/>\n[tex]\\qquad z=\\dfrac{-254+94y}{14}[\/tex]<br \/>\n[tex]\\qquad z=\\dfrac{-127+47y}{7}.\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\textcolor{#800000}{(vi)}[\/tex]<\/p>\n<li>Substituindo [tex]\\textcolor{#800000}{(v)}[\/tex] e [tex]\\textcolor{#800000}{(vi)}[\/tex] em [tex]\\textcolor{#800000}{(i)}[\/tex] temos<\/li>\n<p>[tex]\\qquad 11\\cdot\\bigg(\\dfrac{-127+19y}{7}\\bigg)-7y+2\\cdot\\bigg(\\dfrac{-127+47y}{7}\\bigg)=-127[\/tex]<br \/>\n[tex]\\qquad 11\\cdot(-127+19y)-49y+2\\cdot(-127+47y)=-889[\/tex]<br \/>\n[tex]\\qquad -1397+209y-49y-254+94y=-889[\/tex]<br \/>\n[tex]\\qquad 254y=762[\/tex]<br \/>\n[tex]\\qquad y=\\dfrac{762}{254}[\/tex]<br \/>\n[tex]\\qquad \\boxed{y=3}\\,.\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\textcolor{#800000}{(vii)}[\/tex]<\/p>\n<li>Substituindo [tex]\\textcolor{#800000}{(vii)}[\/tex] em [tex]\\textcolor{#800000}{(v)}[\/tex] e [tex]\\textcolor{#800000}{(vi)}[\/tex], respectivamente, temos que<\/li>\n<p> [tex]\\qquad \\boxed{x=-10}\\,[\/tex] e [tex]\\,\\boxed{z=2}[\/tex]. <\/ul>\n<p>Portanto, [tex]\\fcolorbox{black}{#eee0e5}{$x\\cdot y\\cdot z=(-10)\\cdot 3\\cdot 2=-60$}[\/tex] .<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:10px\">\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 2\u00ba ano do E. M.) Sabendo que [tex](x + 2y\\;,\\; 3x \u2212 5y\\;,\\; 8x \u2212 2y\\;,\\; 11x \u2212 7y + 2z)[\/tex] \u00e9 uma progress\u00e3o aritm\u00e9tica com o \u00faltimo termo igual a [tex]\u2212127[\/tex], calcule [tex]x\\cdot y\\cdot z[\/tex]. (ITA, 2012 &#8211; Adaptado) Lembretes Dada uma progress\u00e3o aritm\u00e9tica [tex](a_{1}, a_{2}, a_{3}, \\dots)[\/tex] de &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-numeros-em-progressao-aritmetica\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":721,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-160243","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/160243","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/721"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=160243"}],"version-history":[{"count":34,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/160243\/revisions"}],"predecessor-version":[{"id":196968,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/160243\/revisions\/196968"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=160243"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}