Question

Find the sum is of the geometric series 12+6+3+...

Answer 1

$S=12+6+3+...\\\\a_1=12;\ a_2=6;\ a_3=3\\\\r=a_2:a_1\\\\r=6:12=\dfrac{1}{2} \ \textless \ 1\\\\therefore\ S=\dfrac{a_1}{1-r}\\\\S=12+6+3+...=\dfrac{12}{1-\frac{1}{2}}=\dfrac{12}{\frac{1}{2}}=12\cdot\dfrac{2}{1}=24\\\\Answer:\boxed{12+6+3+...=24}$