Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 22939 by selestian last updated on 24/Oct/17

Answered by ajfour last updated on 24/Oct/17

$$\left(B\right)1/8$$$$\frac{a}{1-r}=4;\frac{a^3}{1-r^3}=\frac{64}{7}$$$$\Rightarrow \frac{7a^3}{1-r^3}=\frac{a^3}{{(1-r)}^3}$$$$\Rightarrow 7{(1-r)}^2=1+r+r^2$$$$7+7r^2-14r=1+r+r^2$$$$2r^2-5r+2=0$$$$(2r-1)(r-2)=0$$$$\Rightarrow r=\frac{1}{2}forthedecreasingG.P.$$$$\frac{a}{1-r}=4\Rightarrow a=4(1-\frac{1}{2})$$$$a=2,then{5}^{th}term{ar}^4=\frac{1}{8}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com