+129899 y = pq^x
Note that when x = 0, then y = 5
This suggests that
5 = pq^0
5 = p (1)
5 = p
And the point (4, 405) is on the graph..so
405 = 5q^(4) divide both sides by 5
81 = q^4 take the 4th root of each side and we get that
3 = q
So....to find k when x =2, we have
k = 5*(3)^2 = 5 * 9 = 45
+26393 equations simultaneous non-linear
geometric sequence: \(\text{$a_1 = p$, $a_2 = pq$, $a_3 = pq^2$, $a_4=pq^3$, $a_5 = pq^4$, $\ldots$}\)
\(\text{ $k \quad(a_3=pq^2)$ is the geometric mean of $y = 5 \quad(a_1=p)$ and $y = 405 \quad(a_5=pq^4)$ } \\ \huge{ \begin{array}{|rcll|} \hline k&=&\sqrt{5\cdot 405} \\ &=&\sqrt{2025} \\ &=& 45 \\ \hline \end{array} }\)
