Question Number 129892 by mohammad17 last updated on 20/Jan/21
Answered by Ar Brandon last updated on 20/Jan/21
$$\mathrm{GP}\:\mathrm{with}\:\mathrm{U}\left(\mathrm{1}\right)=\frac{\mathrm{11x}}{\mathrm{12}}\:\mathrm{and}\:\mathrm{r}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\mathrm{S}_{\infty} =\frac{\frac{\mathrm{11}}{\mathrm{12}}\mathrm{x}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{12}}}=\mathrm{x} \\ $$