{"id":88932,"date":"2017-11-11T14:09:04","date_gmt":"2017-11-11T16:09:04","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=88932"},"modified":"2023-04-14T02:31:25","modified_gmt":"2023-04-14T05:31:25","slug":"problemao-progressao-geometrica","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-progressao-geometrica\/","title":{"rendered":".Problem\u00e3o:  Progress\u00e3o Geom\u00e9trica"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\" style=\"line-height:30px;\">\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\"\/>\nSejam [tex]a[\/tex], [tex]b[\/tex], [tex]c[\/tex] e [tex]d[\/tex] quatro n\u00fameros em progress\u00e3o geom\u00e9trica, nessa ordem.<br \/>\nProve que [tex](b &#8211; c)^2 + (c &#8211; a)^2 + (d &#8211; b)^2 = (a &#8211; d)^2[\/tex].<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:32px;\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o <\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nSeja [tex]q[\/tex] a raz\u00e3o da progress\u00e3o geom\u00e9trica em quest\u00e3o.<br \/>\nSuponha, inicialmente, que a raz\u00e3o [tex]q[\/tex] e o primeiro termo [tex]a[\/tex] s\u00e3o n\u00e3o nulos; assim, da defini\u00e7\u00e3o de progress\u00e3o geom\u00e9trica temos que:<br \/>\n[tex]\\qquad q = \\dfrac{b}{a} = \\dfrac{c}{b} = \\dfrac{d}{c}[\/tex].<br \/>\nIgualando duas a duas essas fra\u00e7\u00f5es, obtemos [tex]\\boxed{b^2 = ac}[\/tex]; [tex]\\boxed{c^2 = bd}[\/tex] e [tex]\\boxed{bc = ad}. \\quad \\textcolor{#800000}{^{(*)}}[\/tex]<br \/>\nPartindo da express\u00e3o [tex](b &#8211; c)^2 + (c &#8211; a)^2 + (d &#8211; b)^2[\/tex] obtemos:<\/p>\n<p style=\"margin:10px\">\n<p>[tex]\\quad \\begin{align*}(b &#8211; c)^2 + (c &#8211; a)^2 + (d &#8211; b)^2 &#038;= b^2 -2bc + c^2 + c^2 &#8211; 2ac + a^2 + d^2 -2db + b^2\\\\<br \/>\n&#038;= b^2 &#8211; 2ad + c^2 + c^2 &#8211; 2b^2 + a^2 + d^2 &#8211; 2c^2 + b^2\\\\<br \/>\n&#038;= &#8211; 2ad + a^2 + d^2\\\\<br \/>\n&#038;= (a &#8211; d)^2 \\end{align*}[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>[tex]\\textcolor{#800000}{^{(*)}}[\/tex]Observe que essas tr\u00eas igualdades continuam v\u00e1lidas, se [tex]q = 0[\/tex] ou [tex]a = 0[\/tex]; dessa forma, a restri\u00e7\u00e3o inicial de que [tex]q \\, [\/tex] e  [tex] \\, a[\/tex] s\u00e3o n\u00e3o nulos n\u00e3o compromete a generalidade do resultado demonstrado:<br \/>\n [tex]\\qquad \\fcolorbox{black}{#eee0e5}{$(b &#8211; c)^2 + (c &#8211; a)^2 + (d &#8211; b)^2 = (a &#8211; d)^2$}[\/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","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 2\u00ba ano do E. M.) Sejam [tex]a[\/tex], [tex]b[\/tex], [tex]c[\/tex] e [tex]d[\/tex] quatro n\u00fameros em progress\u00e3o geom\u00e9trica, nessa ordem. Prove que [tex](b &#8211; c)^2 + (c &#8211; a)^2 + (d &#8211; b)^2 = (a &#8211; d)^2[\/tex]. Solu\u00e7\u00e3o Seja [tex]q[\/tex] a raz\u00e3o da progress\u00e3o geom\u00e9trica em quest\u00e3o. Suponha, inicialmente, que a raz\u00e3o &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problemao-progressao-geometrica\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":179,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-88932","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/88932","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\/179"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=88932"}],"version-history":[{"count":23,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/88932\/revisions"}],"predecessor-version":[{"id":186893,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/88932\/revisions\/186893"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=88932"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}