What are the first two terms of the sequence defined recursively by the formula a_(n)=(a_(n-1))^2+4 when a_(4)=1604 2,8 6,40 sqrt (2),6 sqrt (6),10

To find the first two terms of the sequence, we need to work backwards from the given term $a_4 = 1604$ using the recursive formula $a_n = (a_{n-1})^2 + 4$.<br /><br />Let's start by finding $a_3$:<br />$a_4 = (a_3)^2 + 4$<br />$1604 = (a_3)^2 + 4$<br />$(a_3)^2 = 1600$<br />$a_3 = \sqrt{1600} = 40$<br /><br />Now, let's find $a_2$:<br />$a_3 = (a_2)^2 + 4$<br />$40 = (a_2)^2 + 4$<br />$(a_2)^2 = 36$<br />$a_2 = \sqrt{36} = 6$<br /><br />Therefore, the first two terms of the sequence are $a_2 = 6$ and $a_3 = 40$. So, the correct answer is:<br /><br />6, 40