Question

Considere a PA ($a_{1}$, $a_{2}$, $a_{3}$, ...) de razão r = $\frac{1}{ \sqrt{2} - 1 }$ e $a_{6}$ = 1 - $\sqrt{2}$. Determine o termo $a_{7}$. 

Answer 1

$r=\frac{1}{\sqrt{2}-1}.\frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{\sqrt{2}+1}{2-1}=\sqrt{2}+1\\\\an=a1+(n-1).r\\a6=1-\sqrt{2}\\1-\sqrt{2}=a1+(6-1).(\sqrt{2}+1)\\1-\sqrt{2}=a1+5.(\sqrt{2}+1)\\1-\sqrt{2}=a1+5\sqrt{2}+5\\1-\sqrt{2}-5\sqrt{2}-5=a1\\a1=-4-6\sqrt{2}\\\\an=a1+(n-1).r\\a7=(-4-6\sqrt{2})+6.(\sqrt{2}+1)\\a7=-4-6\sqrt{2}+6\sqrt{2}+6\\\\\boxed{a7=-4+6=2}$