Question Number 28739 by NECx last updated on 29/Jan/18
$${if}\:{S}_{{n}} =\frac{{a}\left({r}^{{n}} −\mathrm{1}\right)}{{r}−\mathrm{1}}\: \\ $$$$ \\ $$$${make}\:{r}\:{the}\:{subject}\:{of}\:{formula} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by ajfour last updated on 30/Jan/18
$${T}_{{n}} ={ar}^{{n}−\mathrm{1}} \\ $$$$\mathrm{ln}\:\left(\frac{{T}_{{n}} }{{a}}\right)=\left({n}−\mathrm{1}\right)\mathrm{ln}\:{r} \\ $$$$\mathrm{ln}\:{r}\:=\:\frac{\mathrm{1}}{{n}−\mathrm{1}}\mathrm{ln}\:\left(\frac{{S}_{{n}} −{S}_{{n}−\mathrm{1}} }{{a}}\right) \\ $$$${r}=\:{e}^{\frac{\mathrm{1}}{{n}−\mathrm{1}}\mathrm{ln}\:\left(\frac{{S}_{{n}} −{S}_{{n}−\mathrm{1}} }{{a}}\right)} \\ $$$${or}\:\:\boldsymbol{{r}}\:=\:\left(\frac{\boldsymbol{{S}}_{{n}} −\boldsymbol{{S}}_{{n}−\mathrm{1}} }{\boldsymbol{{a}}}\right)^{\frac{\mathrm{1}}{\boldsymbol{{n}}−\mathrm{1}}} \:. \\ $$
Commented by NECx last updated on 30/Jan/18
$${still}\:{having}\:{a}\:{doubt}\:{here}.\:{Does}\:{it} \\ $$$${mean}\:{the}\:{actual}\:{values}\:{cannot} \\ $$$${be}\:{imputed}.\:{If}\:{it}\:{is}\:{done}\:{then}\:{r} \\ $$$${comes}\:{back}\:{to}\:{the}\:{equation}. \\ $$