Question

como calcular razão essa pg a1+a6=1025 e a3*a4=1024

Answer 1

$a_1+a_6=1025\\\\a_1+(a_1\cdot{q}^5)=1025\\\\a_1(q^5+1)=1025\\\\\\\\a_3\cdot{a}_4=1024\\\\(a_1\cdot{q}^2)\cdot(a_1\cdot{q}^3)=1024\\\\a_1^2\cdot{q^5}=1024\\\\q^5=\dfrac{1024}{a_1^2}\\\\\\\\a_1(q^5+1)=1025\\\\\\a_1\left(\left(\dfrac{1024}{a_1^2}\right)+1\right)=1025\\\\\\\dfrac{1024a_1}{a_1^2}+a_1=1025\\\\\\\dfrac{1024}{a_1}+a_1=1025\\\\\\1024+a_1^2=1025a_1\\\\a_1^2-1025a_1+1024=0\\\\(a_1-1)\cdot(a_1-1024)=0\\\\\\\begin{matrix}a_1'-1=0\\\\a_1'=1\end{matrix}\qquad\qquad\qquad\qquad\begin{matrix}a_1''-1024=0\\\\a_1''=1024\end{matrix}$




$\text{Para}\ \ a_1'=1,\text{ temos o seguinte valor de}\ q':\\\\q^{5}=\dfrac{1024}{a_1^2}\\\\\\q^{5}=\dfrac{1024}{(1)^2}\\\\\\q^5=\dfrac{1024}{1}\\\\\\q^5=1024\\\\q=\sqrt[5]{1024}\\\\\boxed{q'=4}\\\\\\\\\text{Para}\ \ a_1''=1024,\text{ temos o seguinte valor de}\ q:\\\\q^5=\dfrac{1024}{a_1^2}\\\\\\q^5=\dfrac{1024}{(1024)^2}\\\\\\q^5=\dfrac{4^5}{(4^5)^2}\\\\\\q^5=\dfrac{4^5}{(4^2)^5}\\\\\\q=\sqrt[5]{\dfrac{4^5}{(4^2)^5}}\\\\\\q=\dfrac{4}{4^2}\\\\\\\boxed{q''=\dfrac{1}{4}}$