[tex]{\huge{\fcolorbox{yellow}{red}{\orange{\boxed{\boxed{\boxed{\boxed{\underbrace{\overbrace{\mathfrak{\pink{\fcolorbox{green}{blue}{Answer}}}}}}}}}}}}}[/tex]
(i)
[tex] \sf{a_n = 20 \times {( \frac{2}{3} )}^{n - 1} }[/tex]
(ii)
[tex] \sf S_n = 60 \{1 - { \frac{2}{3}}^{n} \} [/tex]
Step-by-step explanation:
[tex]\underline\red{\textsf{Given :-}}[/tex]
height of ball (a) = 10m
fraction of height decreases by each bounce (r) = 2/3
[tex] \underline\pink{\textsf{Solution :-}}[/tex]
(i) We will use here geometric progression formula to find height an times
[tex]{\blue{\sf{a_n = a {r}^{n - 1} }}} \\ \sf{a_n = 20 \times { \frac{2}{3} }^{n - 1} }[/tex]
(ii) here we will use the sum formula of geometric progression for finding the total nth impact
[tex] \orange {\sf{S_n = a \times \frac{(1 - {r}^{n} )}{1 - r} }} \\ \sf S_n = 20 \times \frac{1 - ( { \frac{2}{3} })^{n} }{1 - \frac{2}{3} } \\ \sf S_n = 20 \times \frac{1 - {( \frac{2}{3}) }^{n} }{ \frac{1}{3} } \\ \sf S_n = 3 \times 20 \times \{1 - ( { \frac{2}{3}) }^{n} \} \\ \purple{\sf S_n = 60 \{1 - { \frac{2}{3} }^{n} \}}[/tex]
