Question Number 55893 by Tawa1 last updated on 05/Mar/19
$$\mathrm{Three}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{in}\:\mathrm{G}.\mathrm{P}\:\mathrm{such}\:\mathrm{that}\:\mathrm{their}\:\mathrm{sum}\:\mathrm{is}\:\:\boldsymbol{\mathrm{p}}\:\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{square}\:\mathrm{is}\:\:\boldsymbol{\mathrm{q}}.\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{G}.\mathrm{P} \\ $$
Answered by kaivan.ahmadi last updated on 05/Mar/19
$${let}\:{a},{b},{c}\:{are}\:{numbers},\:{so} \\ $$$${b}^{\mathrm{2}} ={ac}. \\ $$$${a}+{b}+{c}={p}\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{ac}+\mathrm{2}{bc}={p}^{\mathrm{2}} \Rightarrow \\ $$$${q}+\mathrm{2}{ab}+\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{bc}={p}^{\mathrm{2}} \Rightarrow{q}+\mathrm{2}{b}\left({a}+{b}+{c}\right)={p}^{\mathrm{2}} \\ $$$${q}+\mathrm{2}{bp}={p}^{\mathrm{2}} \Rightarrow{b}=\frac{{p}^{\mathrm{2}} −{q}}{\mathrm{2}{p}} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 05/Mar/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$