Question Number 45422 by Tawa1 last updated on 12/Oct/18
$$\mathrm{Three}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{G}.\mathrm{P}\:\mathrm{are}\:\mathrm{the}\:\mathrm{3rd},\:\mathrm{5th}\:\mathrm{and}\:\mathrm{8th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{common}\:\mathrm{ratio}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18
$${A}+\left(\mathrm{3}−\mathrm{1}\right){D}={a} \\ $$$${A}+\left(\mathrm{5}−\mathrm{1}\right){D}={ar} \\ $$$${A}+\left(\mathrm{8}−\mathrm{1}\right){D}={ar}^{\mathrm{2}} \\ $$$$\left\{{A}+\mathrm{4}{D}\right\}−\left\{{A}+\mathrm{2}{D}\right\}={ar}−{a} \\ $$$$\mathrm{2}{D}={ar}−{a} \\ $$$${D}=\frac{{ar}−{a}}{\mathrm{2}} \\ $$$${A}+\mathrm{2}{D}={a} \\ $$$${A}+{ar}−{a}={a} \\ $$$${A}=\mathrm{2}{a}−{ar} \\ $$$${A}+\mathrm{7}{D}={ar}^{\mathrm{2}} \\ $$$$\mathrm{2}{a}−{ar}+\mathrm{7}\left(\frac{{ar}−{a}}{\mathrm{2}}\right)={ar}^{\mathrm{2}} \\ $$$$\mathrm{4}{a}−\mathrm{2}{ar}+\mathrm{7}{ar}−\mathrm{7}{a}=\mathrm{2}{ar}^{\mathrm{2}} \\ $$$$\mathrm{4}−\mathrm{2}{r}+\mathrm{7}{r}−\mathrm{7}−\mathrm{2}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$−\mathrm{2}{r}^{\mathrm{2}} +\mathrm{5}{r}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{2}{r}^{\mathrm{2}} −\mathrm{5}{r}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{2}{r}^{\mathrm{2}} −\mathrm{2}{r}−\mathrm{3}{r}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{2}{r}\left({r}−\mathrm{1}\right)−\mathrm{3}\left({r}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({r}−\mathrm{1}\right)\left(\mathrm{2}{r}−\mathrm{3}\right)=\mathrm{0} \\ $$$${r}=\mathrm{1}\:\:\:{or}\:{r}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by Tawa1 last updated on 12/Oct/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18
$${God}\:{bless}\:{all} \\ $$