The problem asks us to find the 5th term ($U_5$) of a geometric sequence, given that:

• The first term ($a$) is 5,
• The third term ($U_3$) is 45,
• The common ratio ($r$) is positive.

In a geometric sequence, the $n$-th term is given by the formula: $U_n = a \cdot r^{n-1}$ For the third term ($U_3$): $U_3 = a \cdot r^2$ Substituting the known values: $45 = 5 \cdot r^2$ Dividing both sides by 5: $r^2 = 9$ Taking the positive square root (since $r$ is positive): $r = 3$

Now, we calculate the 5th term ($U_5$): $U_5 = a \cdot r^4 = 5 \cdot 3^4 = 5 \cdot 81 = 405$ Multiplying by 3: [ U_5 = 5r_