The number of bacteria in a certain culture doubles every hour. If there were $30$ bacteria present in the culture originally, how many bacteria will be present at the end of $2^{\text {nd }}$ hour, $4^{\text {th }}$ hour and $n^{\text {th }}$ hour $?$
It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a $G.P.$
Here, $a=30$ and $r=2 \quad \therefore a_{3}=a r^{2}=(30)(2)^{2}=120$
Therefore, the number of bacteria at the end of $2^{\text {nd }}$ hour will be $120 .$
$a_{5}=a r^{4}=(30)(2)^{4}=480$
The number of bacteria at the end of $4^{\text {th }}$ hour will be $480 . $
$a_{n+1}=a r^{n}=(30) 2^{n}$
Thus, number of bacteria at the end of $n^{t h}$ hour will be $30(2)^{n}$
If $a$,$b$,$c \in {R^ + }$ are such that $2a$,$b$ and $4c$ are in $A$.$P$ and $c$,$a$ and $b$ are in $G$.$P$., then
The sum of a $G.P.$ with common ratio $3$ is $364$, and last term is $243$, then the number of terms is
Sum of infinite number of terms in $G.P.$ is $20$ and sum of their square is $100$. The common ratio of $G.P.$ is
The solution of the equation $1 + a + {a^2} + {a^3} + ....... + {a^x}$ $ = (1 + a)(1 + {a^2})(1 + {a^4})$ is given by $x$ is equal to
The value of $0.\mathop {234}\limits^{\,\,\, \bullet \,\, \bullet } $ is