@import url(https://fonts.googleapis.com/css?family=Source+Sans+Pro:300,400,600,700); Questão 55- Seja \ufeff(a1,a2,a3,a4,a5,\u22ef\u2009)(a_1, a_2, a_3, a_4, a_5, \cdots) (a1\u200b,a2\u200b,a3\u200b,a4\u200b,a5\u200b,\u22ef)\ufeff uma PG de termos não nulos. Se \ufeff2(a2+a4)=a3+a52(a_2 + a_4) = a_3 + a_52(a2\u200b+a4\u200b)=a3\u200b+a5\u200b\ufeff , pode-se afirmar corretamente que a razão dessa PG éa) 4 b) 2 c) \ufeff12\frac{1}{2}21\u200b\ufeff d) \ufeff2\sqrt{2}2\u200b\ufeffRESOLUÇÃO Fala galera, para resolvermos essa questão, precisamos lembrar o que é uma P.G, é toda sequência numérica onde todo termo é igual ao produto de seu antecessor com uma constante chamada razão da PG.\ufeffa1=a1a_1 = a_1a1\u200b=a1\u200b\ufeff\ufeffa2=a1\u22c5qa_2=a_1\cdot q a2\u200b=a1\u200b\u22c5q\ufeff\ufeffa3=a1\u22c5q2a_3=a_1\cdot q ^2a3\u200b=a1\u200b\u22c5q2\ufeff\ufeffa4=a1\u22c5q3a_4=a_1\cdot q ^3a4\u200b=a1\u200b\u22c5q3\ufeff\ufeffa5=a1\u22c5q4a_5=a_1\cdot q ^4a5\u200b=a1\u200b\u22c5q4\ufeff\ufeff\ufeff2(a1\u22c5q+a1\u22c5q3)=a1\u22c5q2+a1\u22c5q4\ufeff2(a_1 \cdot q + a_1 \cdot q^3) = a_1 \cdot q ^2+ a_1 \cdot q^4\ufeff2(a1\u200b\u22c5q+a1\u200b\u22c5q3)=a1\u200b\u22c5q2+a1\u200b\u22c5q4\ufeff\ufeff\ufeff2a1\u22c5q+2a1\u22c5q3=a1\u22c5q2+a1\u22c5q4\ufeff2a_1 \cdot q + 2a_1 \cdot q^3 = a_1 \cdot q ^2+ a_1 \cdot q^4\ufeff2a1\u200b\u22c5q+2a1\u200b\u22c5q3=a1\u200b\u22c5q2+a1\u200b\u22c5q4\ufeff\ufeff\ufeff\ufeffa1(2\u22c5q+2\u22c5q3)=a1(q2+q4)\ufeff\ufeff\ufeffa_1( 2\cdot q + 2\cdot q^3) = a_1 (q ^2+ q^4)\ufeff\ufeff\ufeffa1\u200b(2\u22c5q+2\u22c5q3)=a1\u200b(q2+q4)\ufeff\ufeff\ufeff\ufeff2\u22c5q+2\u22c5q3=q2+q4\ufeff2 \cdot q + 2 \cdot q^3= q^2 + q^4\ufeff2\u22c5q+2\u22c5q3=q2+q4\ufeff\ufeff2q(1+q2)=q2(1+q2)2q( 1 + q^2)= q^2(1+q^2)2q(1+q2)=q2(1+q2)\ufeff\ufeff2q=q22q=q^22q=q2\ufeff\ufeff2=q2=q2=q\ufeffGABARITO LETRA: (B)Qualquer duvida, meu instagram @carol.1111
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