Question

determine a razão da progressão geométrica (a+ 1,a+4,a+10)​

Answer 1

Se a sequencia apresentada é uma PG, então sua razão deve ser mantida constante, ou seja:

$razao:~~~\boxed{\dfrac{a_3}{a_2}~=~\dfrac{a_2}{a_1}}\\\\\\Substituindo~os~termos~na~equcao:\\\\\\\dfrac{a+10}{a+4}~=~\dfrac{a+4}{a+1}\\\\\\Multiplicando~cruzado\\\\\\(a+10)~.~(a+1)~=~(a+4)~.~(a+4)\\\\\\a^2+10a+a+10~=~a^2+4a+4a+16\\\\\\a^2-a^2+11a-8a~=~16-10\\\\\\3a~=~6\\\\\\a~=~\dfrac{6}{2}\\\\\\\boxed{a~=~3}$

Com o valor de "a", podemos agora determinar a razão:

$razao~=~\dfrac{a_2}{a_1}\\\\\\razao~=~\dfrac{a+4}{a+1}\\\\\\razao~=~\dfrac{2+4}{2+1}\\\\\\razao~=~\dfrac{6}{3}\\\\\\\boxed{razao~=~2}$