Given, the quantity of bacteria copies each hour. Henceforth, the quantity of bacteria after consistently will frame a G.P.
Here we have, \[a\text{ }=\text{ }30\text{ }and\text{ }r\text{ }=\text{ }2\]
Thus, \[{{a}_{3}}~=~a{{r}^{2}}~=\text{ }\left( 30 \right)\text{ }{{\left( 2 \right)}^{2}}~=\text{ }120\]
In this manner, the quantity of bacteria toward the finish of \[{{2}^{nd}}~hour\text{ }will\text{ }be\text{ }120.\]
Also, \[{{a}_{5}}~=~a{{r}^{4}}~=\text{ }\left( 30 \right)\text{ }{{\left( 2 \right)}^{4}}~=\text{ }480\]
The quantity of bacteria toward the finish of \[{{4}^{th}}~hour\text{ }will\text{ }be\text{ }480.\]
\[{{a}_{n}}_{~+1~}=~a{{r}^{n}}~=\text{ }\left( 30 \right)\text{ }{{2}^{n}}\]
In this manner, the quantity of bacteria toward the finish of\[{{n}^{th}}~hour\text{ }will\text{ }be\text{ }30{{\left( 2 \right)}^{n}}.\]