Xavier Aptitude Test 2022 Paper

Given the ball falls from a height of $3\mathrm{meters}$.
The ball reaches a height which $0.8\mathrm{times}$ the original height every time.
Hence this is in the form of a geometric progression. We need to count distance when the ball flies upward and downward.
Hence considering every time the ball flies upward to a series with terms :
${\mathit{h}}_1,{\mathit{h}}_2,........$
Every time the ball falls down to be:
${\mathit{d}}_1,{\mathit{d}}_2,.......$
${\mathit{h}}_1=\left(0.8\right)\times 3,{\mathit{h}}_2=\left(0.8\right)\times \left(0.8\right)\times 3,..........$
${\mathit{d}}_1=3,{\mathit{d}}_2=3\times \left(0.8\right),{\mathit{d}}_3=3\times \left(0.8\right)\times \left(0.8\right)........$
${\mathit{h}}_1+{\mathit{h}}_2........=$Sum of an infinite geometric progression. $=3\times 0.8\left(1+0.8+0.64+......\right)$
The sum of an infinite $\mathrm{GP}$ with $\mathit{r}$ less than $1$ is : $\frac{\mathit{a}}{1-\mathit{r}}$
$=2.4.\left(\frac{1}{1-0.8}\right)=12\mathrm{meters}$
The sum of ${\mathit{d}}_1+{\mathit{d}}_2+.........$
$=3+\left({\mathit{h}}_1+{\mathit{h}}_2+......\right)=15$
The total distance $=15+12=27\mathrm{meters}$