, 2022

If $r$ is a term in the sequence and $s$ is the next term, then $s = 1 + \dfrac{1}{1+r}$. This means that $s - 1 =
$\dfrac{1}{1+r}$$and so$$\dfrac{1}{s-1}$ = 1+r$which gives$r = $\dfrac{1}{s-1}$ - 1$. Therefore, since$a_3 =
$\dfrac{41}{29}$$, then$$a_2 = \dfrac{1}{a_3-1} - 1 = \dfrac{1}{(41/29)-1} - 1 = \dfrac{1}{12/29} - 1 = \dfrac{29}{12} - 1 = \dfrac{17}{12}$$Further, since$a_2 = $\dfrac{17}{12}$$, then$$a_1 = \dfrac{1}{a_2-1} - 1 = \dfrac{1}{(17/12)-1} - 1 = \dfrac{1}{5/12} - 1 = \dfrac{12}{5} - 1 = \dfrac{7}{5}$ Initially, the water in the hollow tube forms a cylinder with radius 10 mm and height

h mm. Thus, the volume of the water is

$\text{(10 mm ) 2(h}$ mm}) = $\text{100 h}$ mm}^3. After the rod is inserted, the level of the water rises to 64 mm. Note that this does not overflow the tube, since the tube’s height is 100 mm. Up to the height of the water, the tube is a cylinder with radius 10 mm and height 64 mm. Thus, the volume of the tube up to the height of the water is

$\text{(10 mm ) 2(64 mm ) = 6400 mm 3}$ This volume consists of the water that is in the tube (whose volume, which has not changed, is

$\text{100 h}$ mm}^3) and the rod up to a height of 64 mm. [[IMAGE0]] Since the radius of the rod is 2.5 mm, the volume of the rod up to a height of 64 mm is

$\text{(2.5 mm ) 2(64}$ mm}) = $\text{400}$ mm}^3$. Comparing volumes,$$\text{6400}$ mm}^3 =
$\text{100 h}$ mm}^3 + $\text{400}$ mm}^3$and so$100h = 6000$which gives$h = 60$.