Question Number 210629 by peter frank last updated on 14/Aug/24
Answered by Rasheed.Sindhi last updated on 14/Aug/24
$$\begin{cases}{{x}={a}+\left({a}+{d}\right)+\left({a}+\mathrm{2}{d}\right)+,…+\left({a}+\left({m}−\mathrm{1}\right){d}\right)}\\{{y}=\left({a}+{md}\right)+\left({a}+\left({m}+\mathrm{1}\right){d}\right)+…+\left({a}+\left(\mathrm{2}{m}−\mathrm{1}\right){d}\right)}\\{{z}=\left({a}+\mathrm{2}{md}\right)+\left({a}+\left(\mathrm{2}{m}+\mathrm{1}\right){d}\right)+…+\left({a}+\left(\mathrm{3}{m}−\mathrm{1}\right){d}\right)}\end{cases}\: \\ $$$$\begin{cases}{{x}−{ma}=\left({d}\right)+\left(\mathrm{2}{d}\right)+,…+\left(\left({m}−\mathrm{1}\right){d}\right)}\\{{y}−{ma}=\left({md}\right)+\left(\left({m}+\mathrm{1}\right){d}\right)+…+\left(\left(\mathrm{2}{m}−\mathrm{1}\right){d}\right)}\\{{z}−{ma}=\left(\mathrm{2}{md}\right)+\left(\left(\mathrm{2}{m}+\mathrm{1}\right){d}\right)+…+\left(\left(\mathrm{3}{m}−\mathrm{1}\right){d}\right)}\end{cases}\:\: \\ $$$$\begin{cases}{\frac{{x}−{ma}}{{d}}=\left(\mathrm{1}\right)+\left(\mathrm{2}\right)+,…+\left(\left({m}−\mathrm{1}\right)\right)}\\{\frac{{y}−{ma}}{{d}}=\left({m}\right)+\left(\left({m}+\mathrm{1}\right)\right)+…+\left(\left(\mathrm{2}{m}−\mathrm{1}\right)\right)\:}\\{\frac{{z}−{ma}}{{d}}=\left(\mathrm{2}{m}\right)+\left(\left(\mathrm{2}{m}+\mathrm{1}\right)\right)+…+\left(\left(\mathrm{3}{m}−\mathrm{1}\right)\right)}\end{cases}\: \\ $$$$…\:\: \\ $$
Answered by mr W last updated on 14/Aug/24
$${say}\:{the}\:{difference}\:{from}\:{the}\:\left({m}+\mathrm{1}\right)^{{th}} \\ $$$${and}\:{the}\:{first}\:{term}\:{is}\:{t},\:{then} \\ $$$${each}\:{term}\:{in}\:{the}\:{second}\:{group}\:{is} \\ $$$${by}\:{t}\:{larger}\:{than}\:{the}\:{corresponding} \\ $$$${term}\:{in}\:{the}\:{first}\:{group}.\:{the}\:{sum}\:{of} \\ $$$${the}\:{second}\:{group}\:{is}\:{by}\:{mt}\:{larger}\:{than} \\ $$$${the}\:{sum}\:{of}\:{first}\:{group},\:{etc}. \\ $$$${a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:…,\:{a}_{{m}} \:\:\leftarrow\:{the}\:{first}\:{m}\:{terms},\:{their}\:{sum}\:{is}\:{x} \\ $$$${b}_{\mathrm{1}} ,\:{b}_{\mathrm{2}} ,\:…,\:{b}_{{m}} \:\:\leftarrow\:{the}\:{next}\:{m}\:{terms},\:{their}\:{sum}\:{is}\:{y} \\ $$$${c}_{\mathrm{1}} ,\:{c}_{\mathrm{2}} ,\:…,\:{c}_{{m}} \:\:\leftarrow\:{the}\:{last}\:{m}\:{terms},\:{their}\:{sum}\:{is}\:{z} \\ $$$${b}_{\mathrm{1}} ={a}_{\mathrm{1}} +{t},\:{b}_{\mathrm{2}} ={a}_{\mathrm{2}} +{t},\:…,\:{b}_{{m}} ={a}_{{m}} +{t} \\ $$$${c}_{\mathrm{1}} ={b}_{\mathrm{1}} +{t},\:{c}_{\mathrm{2}} ={b}_{\mathrm{2}} +{t},\:…,\:{c}_{{m}} ={b}_{{m}} +{t} \\ $$$${y}={x}+{mt} \\ $$$${z}={y}+{mt} \\ $$$$\Rightarrow{x}+{z}=\mathrm{2}{y}\: \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}{xz}=\mathrm{4}{y}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{xz}=\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{xz} \\ $$$$\Rightarrow\left({x}−{z}\right)^{\mathrm{2}} =\mathrm{4}\left({y}^{\mathrm{2}} −{xz}\right)\:\checkmark \\ $$
Commented by mm1342 last updated on 14/Aug/24
$${very}\:\:{excellent}\:\:\:\:\:\underline{\underbrace{\lesseqgtr}} \\ $$