Question Number 68775 by rajesh4661kumar@gmail.com last updated on 15/Sep/19
![](https://www.tinkutara.com/question/9312.png)
Answered by $@ty@m123 last updated on 15/Sep/19
![Let a be the first term and d be the common difference of AP. a+(p−1)d=A ...(1) a+(q−1)d=AR ....(2) a+(r−1)d=AR^2 ....(3) a+(s−1)d=AR^3 ....(4) (1)−(2) (p−q)d=A−AR p−q=((A(1−R))/d) ...(5) q−r=((AR(1−R))/d) ...(6) r−s=((AR^2 (1−R))/d) ...(7) We observe that: ((q−r)/(p−q))=((r−s)/(q−r))=R Hence, p−q, q−r,r−s are in G.P.](https://www.tinkutara.com/question/Q68797.png)
$${Let}\:{a}\:{be}\:\:{the}\:{first}\:{term}\:{and}\:{d}\:{be}\:{the} \\ $$$${common}\:{difference}\:{of}\:{AP}. \\ $$$${a}+\left({p}−\mathrm{1}\right){d}={A}\:…\left(\mathrm{1}\right) \\ $$$${a}+\left({q}−\mathrm{1}\right){d}={AR}\:\:….\left(\mathrm{2}\right) \\ $$$${a}+\left({r}−\mathrm{1}\right){d}={AR}^{\mathrm{2}} \:….\left(\mathrm{3}\right) \\ $$$${a}+\left({s}−\mathrm{1}\right){d}={AR}^{\mathrm{3}} \:….\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right) \\ $$$$\left({p}−{q}\right){d}={A}−{AR} \\ $$$${p}−{q}=\frac{{A}\left(\mathrm{1}−{R}\right)}{{d}}\:\:…\left(\mathrm{5}\right) \\ $$$${q}−{r}=\frac{{AR}\left(\mathrm{1}−{R}\right)}{{d}}\:\:…\left(\mathrm{6}\right) \\ $$$${r}−{s}=\frac{{AR}^{\mathrm{2}} \left(\mathrm{1}−{R}\right)}{{d}}\:\:…\left(\mathrm{7}\right) \\ $$$${We}\:{observe}\:{that}: \\ $$$$\frac{{q}−{r}}{{p}−{q}}=\frac{{r}−{s}}{{q}−{r}}={R} \\ $$$${Hence}, \\ $$$${p}−{q},\:{q}−{r},{r}−{s}\:{are}\:{in}\:{G}.{P}. \\ $$
Commented by peter frank last updated on 15/Sep/19
![thank you](https://www.tinkutara.com/question/Q68803.png)
$${thank}\:\:{you} \\ $$