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Question Number 99603 by bemath last updated on 22/Jun/20

$$\mathrm{How}\mathrm{many}\mathrm{terms}\mathrm{in}\mathrm{the}\mathrm{geometric}$$$$\mathrm{progression}1,1.1,1.21,1.331,...$$$$\mathrm{will}\mathrm{be}\mathrm{needed}\mathrm{so}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}$$$$\mathrm{the}\mathrm{first}n\mathrm{terms}\mathrm{is}\mathrm{greather}\mathrm{than}20?$$

Commented by john santu last updated on 22/Jun/20

$$\mathrm{The}\mathrm{sequence}\mathrm{is}\mathrm{a}\mathrm{geometric}\mathrm{progression}$$$$\mathrm{with}\{\begin{array}{l}a=1\\ r=1.1\end{array}$$$$\mathrm{we}\mathrm{want}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{smallest}\mathrm{value}\mathrm{of}$$$$n\mathrm{such}\mathrm{that}{S}_n>20$$$$\Rightarrow {\mathrm{S}}_n=\frac{a(1-r^n)}{1-r}=\frac{1.(1-{(1.1)}^n}{1-1.1}>20$$$$\frac{1-{(1.1)}^n}{-0.1}>20;{(1.1)}^n-1>2$$$${(1.1)}^n>3;\mathrm{n}>\frac{\ln 3}{\ln 1.1};n>11.526$$$$\mathrm{and}\mathrm{therefore}\mathrm{the}\mathrm{smallest}\mathrm{number}$$$$\mathrm{value}\mathrm{of}n\mathrm{is}12◼$$

Commented by bemath last updated on 22/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{both}$$

Answered by Rio Michael last updated on 22/Jun/20

$$S_n=\frac{a(r^n-1)}{r-1}$$$$S_n=\frac{{(1.1)}^n-1}{1.1-1}=\frac{{(1.1)}^n-1}{0.1}$$$$\mathrm{for}{S}_n>20,\Rightarrow \frac{{(1.1)}^n-1}{0.1}>20\Rightarrow {(1.1)}^n>3$$$$n\log (1.1)>\log 3$$$$\Rightarrow n>11.52,\mathrm{but}n\in \mathbb{Z}\Rightarrow n=12.$$

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