Question Number 11633 by Joel576 last updated on 29/Mar/17
$$\mathrm{A}\:\mathrm{geometric}\:\mathrm{sequence}\:\mathrm{with}\:{n}\:\mathrm{terms}\: \\ $$$${a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} ,\:…,\:{a}_{{n}} \:\mathrm{which}\:\mathrm{has}\:{a}_{\mathrm{1}} \:.\:{a}_{{n}} \:=\:\mathrm{3} \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{all}\:{n}\:\mathrm{terms}\:=\:{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} …{a}_{{n}} =\:\mathrm{59049} \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{n} \\ $$
Answered by linkelly0615 last updated on 29/Mar/17
$$ \\ $$$$… \\ $$$${Set}:\:{a}_{{k}} ={a}_{\mathrm{1}} \centerdot{r}^{{k}−\mathrm{1}} \\ $$$$\left({Because}\:{the}\:{sequence}\:{is}\:{a}\:{geometric}\:{sequence}.\right) \\ $$$$\Rightarrow{a}_{\mathrm{1}} \centerdot{a}_{{n}} ={a}_{\mathrm{1}} \centerdot\left({a}_{\mathrm{1}} \centerdot{r}^{{n}−\mathrm{1}} \right)={a}_{\mathrm{1}} ^{\mathrm{2}} \centerdot{r}^{{n}−\mathrm{1}} \\ $$$$\:\because{a}_{{k}} ={a}_{\mathrm{1}} \centerdot{r}^{{k}−\mathrm{1}} \\ $$$$\therefore{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} …{a}_{{n}} ={a}_{\mathrm{1}} \left({a}_{\mathrm{1}} {r}\right)\left({a}_{\mathrm{1}} {r}^{\mathrm{2}} \right)…\left({a}_{\mathrm{1}} {r}^{{n}−\mathrm{1}} \right) \\ $$$$={a}_{\mathrm{1}} ^{{n}} {r}^{\left(\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\right)} =\left({a}_{\mathrm{1}} ^{\mathrm{2}} {r}^{{n}−\mathrm{1}} \right)^{\left(\frac{{n}}{\mathrm{2}}\right)} \\ $$$$=\mathrm{3}^{\frac{{n}}{\mathrm{2}}} =\mathrm{59049} \\ $$$$\Rightarrow\frac{{n}}{\mathrm{2}}=\mathrm{log}\:_{\mathrm{3}} \mathrm{59049}=\mathrm{10} \\ $$$$\left({That}\:{means}\:\mathrm{3}^{\mathrm{10}} =\mathrm{59049}\right) \\ $$$$\Rightarrow\boldsymbol{{n}}=\mathrm{20} \\ $$
Commented by Joel576 last updated on 29/Mar/17
$${thank}\:{you}\:{very}\:{much} \\ $$