Answer: C. [tex]\sqrt2,\ 6[/tex]
Step-by-step explanation:
Given formula [tex]a_n=(a_{n-1})^2+4[/tex]
when[tex]a_4=1604[/tex]
Substitute n=4 in th given formula , we get
[tex]a_4=(a_{4-1})^2+4\\\Rightarrow1604=(a_3)^2+4\\\Rightarrow(a_3)^2=1604-4\\\Rightarrow(a_3)^2=1600\\\Rightarrow\ a_3=40[/tex]
Substitute n=3 in the given formula, we get
[tex]a_3=(a_{3-1})^2+4\\\Rightarrow40=(a_2)^2+4\\\Rightarrow(a_2)^2=40-4\\\Rightarrow(a_2)^2=36\\\Rightarrow\ a_2=6[/tex]
Substitute n=2 in the given formula, we get
[tex]a_2=(a_{2-1})^2+4\\\Rightarrow6=(a_1)^2+4\\\Rightarrow(a_1)^2=6-4\\\Rightarrow(a_1)^2=2\\\Rightarrow\ a_1=\sqrt{2}[/tex]
Thus, the first two terms of the sequence= [tex]\sqrt{2}\ and \ 6[/tex]
