Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 79340 by TawaTawa last updated on 24/Jan/20

Commented by john santu last updated on 24/Jan/20

$$letthisnumber:a,ar,{ar}^2$$$$\left(i\right)a\left(\frac{r^3-1}{r-1}\right)=p\Rightarrow \frac{r^3-1}{r-1}=\frac{p}{a}$$$$r^2+r+1=\frac{p}{a}$$$$\left(ii\right)a^2+a^2{r}^2+a^2{r}^4=q$$$$a^2\left(\frac{r^6-1}{r^2-1}\right)=q\Rightarrow a^2\left(\frac{(r^3-1)(r^3+1)}{(r-1)(r+1)}\right)=q$$$$a^2\left(\frac{p}{a}\right)\left(\frac{r^3+1}{r+1}\right)=q$$$$ap(r^2-r+1)=q$$$$r^2-r+1=\frac{q}{ap}$$$$\left(i\right)-\left(ii\right)2r=\frac{p}{a}-\frac{q}{ap}$$$$r=\frac{p^2-q}{2ap}.sothemidlleterm$$$$isequalto:a\times \left(\frac{p^2-q}{2ap}\right)=\frac{p^2-q}{2p}$$

Commented by TawaTawa last updated on 24/Jan/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir},\mathrm{i}\mathrm{appreciate}$$

Commented by john santu last updated on 24/Jan/20

$$thanksyou$$

Answered by john santu last updated on 24/Jan/20

$$ansC$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com