The half-life of a drug is 3.0 hrs. A patient is administered a 5 mg dose of this drug at an  interval of 6.0 hrs. Calculate the mass (in mg ) of drug remaining in the patient’s body  just  after  sixth dose. Drug follows first order kinetics.

Ans: 5.64 to 5.67
SOL- Between two successive dose, the former dose spent two-half lives.
$\Rightarrow$  At the time when sixth dose is given 
First dose has spent 30 hrs=10 half-lives
Second dose has spend 24 hrs= 8 half-lives
Third dose has spent 18 hrs=6 half-lives
Fourth dose has spent 12 hrs=4 half-lives
Fifth dose has spent 6 hrs= 2 half-lives
Sixth dose is just given =present completely.
$\Rightarrow$  mass of drug remaining 
 $=5[1+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^4+{\left(\frac{1}{2}\right)}^6+{\left(\frac{1}{2}\right)}^8+{\left(\frac{1}{2}\right)}^{10}]$
The above is a geometric progression series with common ratio $=\frac{1}{4}$
 $\Rightarrow$  sum  $=\frac{1-r^n}{1-r}=\frac{1-{\left(\frac{1}{4}\right)}^6}{1-\frac{1}{4}}=1.33$
$\Rightarrow$  Mass of the drug remaining  $=5\times 1.33=6.65mg$