If a rubber ball is dropped from a height of $1\,\mathrm{m}$ and continues to rebound to a height that is nine tenth of its previous fall, find the total distance in meter that it travels on falls only.
I tried if it could be solved using arithmetic progression for which the first term is $(a) = 1\,\mathrm{m}$ and the common difference is $(d) = \frac{9}{10}$. But I could not get any more information.
On
The ball follows geometric progression, because, after each fall, the ball goes back up $\frac 9{10}$th of the height of the previous fall.
That is, if the ball fell down from $x$ metres, the next fall would be from $\frac 9{10} x$, the next from $\frac 9{10} \cdot \frac 9{10} x$.
Since the initial height is $1$ metre and we only have to evaluate the distance the ball falls, the excercise reduces to the sum of a simple Geometric progression, given by $$ 1+ \frac 9{10} + \frac {9^2}{10^2} + \frac {9^3}{10^3} + \cdots \infty$$
$$= \frac 1{1-\frac 9{10}}$$ $$= 10$$ The total distance covered by the ball is $10$ metres.
During the first drop it covers a distance of $1$m and then rises to a distance of $1\bullet\frac{9}{10}$ and then falls to that distance again to rise by $1\bullet\frac{9}{10}\bullet\frac{9}{10}$ and so on. You will notice that this is forming a geometric progression than a arithmetic one.
The summation of all the distances will be,
$1+1\bullet\frac{9}{10}+1\bullet(\frac{9}{10})^2+1\bullet(\frac{9}{10})^3...$
$=\frac{1}{1-\frac{9}{10}}=10$