Question Number 21276 by ketto last updated on 18/Sep/17
Answered by $@ty@m last updated on 19/Sep/17
$$\left({a}\right)\:{Given} \\ $$$${t}_{\mathrm{3}} =\mathrm{18}\:\Rightarrow\:{ar}^{\mathrm{2}} =\mathrm{18}\:\:\:\:\:\:\:−−\left(\mathrm{1}\right) \\ $$$${t}_{\mathrm{5}} =\mathrm{162}\:\Rightarrow\:{ar}^{\mathrm{4}} =\mathrm{162}\:\:−−\left(\mathrm{2}\right) \\ $$$${Dividing}\:\left(\mathrm{2}\right)\:{by}\:\left(\mathrm{1}\right) \\ $$$${r}^{\mathrm{2}} =\mathrm{9}\:\Rightarrow{r}=\pm\mathrm{3} \\ $$$$\therefore\:{from}\:\left(\mathrm{1}\right) \\ $$$${a}=\mathrm{2} \\ $$$$\left({b}\right)\:{Given} \\ $$$${t}_{\mathrm{8}} ={t}_{\mathrm{5}} +\mathrm{9}\:\Rightarrow{a}+\mathrm{7}{d}={a}+\mathrm{4}{d}+\mathrm{9}\Rightarrow{d}=\mathrm{3} \\ $$$${and}\:{t}_{\mathrm{10}} =\mathrm{10}{t}_{\mathrm{2}} \Rightarrow{a}+\mathrm{9}{d}=\mathrm{10}\left({a}+{d}\right) \\ $$$$\Rightarrow{a}+\mathrm{9}{d}=\mathrm{10}{a}+\mathrm{10}{d} \\ $$$$\Rightarrow−\mathrm{9}{a}={d} \\ $$$$\Rightarrow−\mathrm{9}{a}=\mathrm{3} \\ $$$$\Rightarrow{a}=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$