Question Number 81731 by zainal tanjung last updated on 15/Feb/20
$$\mathrm{If}\:\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP};\:\:{p},\:{q},\:{r}\:\mathrm{are}\:\mathrm{in}\:\mathrm{HP}\: \\ $$$$\mathrm{and}\:\:{ap},\:{bq},\:{cr}\:\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP},\:\mathrm{then}\:\frac{{p}}{{r}}+\frac{{r}}{{p}} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
Commented by jagoll last updated on 15/Feb/20
$$\left({i}\right)\:{a},{b},{c}\:{in}\:{AP}\:\Rightarrow\:\mathrm{2}{b}\:=\:{a}+{c} \\ $$$$\left({ii}\right)\:{ap},\:{bq},\:{cr}\:{in}\:{GP}\:\Rightarrow\:\left({bq}\right)^{\mathrm{2}} ={apcr} \\ $$$$\left({iii}\right){p},\:{q},\:{r}\:{in}\:{HP}\:\Rightarrow\:\frac{\mathrm{1}}{{q}}\:=\:\frac{\mathrm{2}}{\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{r}}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{r}}\:=\:\mathrm{2}{q} \\ $$