Let from the height at which ball was thrown = h units
It is given that , ball rebounds the same percentage on each bounce.
Let it rebounds by k % after each bounce.
Height that ball attains after thrown from height h(on 1 st bounce)= ![h + \frac{h k}{100}=h \times (1+\frac{k}{100})](https://tex.z-dn.net/?f=h%20%2B%20%5Cfrac%7Bh%20k%7D%7B100%7D%3Dh%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29)
Height that ball attains after thrown from height h (on 2 n d bounce)= ![h \times (1+\frac{k}{100})+h \times (1+\frac{k}{100})\times \frac{k}{100}=h \times (1+\frac{k}{100})^2](https://tex.z-dn.net/?f=h%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29%2Bh%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29%5Ctimes%20%5Cfrac%7Bk%7D%7B100%7D%3Dh%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29%5E2)
Similarly, the pattern will form geometric sequence.
S= ![h +h \times (1+\frac{k}{100})+h \times (1+\frac{k}{100})^2+h \times (1+\frac{k}{100})^3+.........](https://tex.z-dn.net/?f=h%20%2Bh%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29%2Bh%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29%5E2%2Bh%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29%5E3%2B.........)
So, Common Ratio = ![\frac{\text{2nd term}}{\text{1 st term}}=1 +\frac{k}{100}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B2nd%20term%7D%7D%7B%5Ctext%7B1%20st%20term%7D%7D%3D1%20%2B%5Cfrac%7Bk%7D%7B100%7D)
Common Ratio= 1 + the percentage by which ball rebounds after each bounce
the percentage by which ball rebounds after each bounce= negative integer= k is negative integer.
A is equal to 5
B is equal to 12.
Answer:
Step-by-step explanation:
12 boxes of hay
The second question's answer is 1