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Question-210629

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Question Number 210629 by peter frank last updated on 14/Aug/24
Answered by Rasheed.Sindhi last updated on 14/Aug/24
{ ((x=a+(a+d)+(a+2d)+,...+(a+(m−1)d))),((y=(a+md)+(a+(m+1)d)+...+(a+(2m−1)d))),((z=(a+2md)+(a+(2m+1)d)+...+(a+(3m−1)d))) :} { ((x−ma=(d)+(2d)+,...+((m−1)d))),((y−ma=(md)+((m+1)d)+...+((2m−1)d))),((z−ma=(2md)+((2m+1)d)+...+((3m−1)d))) :} { ((((x−ma)/d)=(1)+(2)+,...+((m−1)))),((((y−ma)/d)=(m)+((m+1))+...+((2m−1)) )),((((z−ma)/d)=(2m)+((2m+1))+...+((3m−1)))) :} ...
$$\begin{cases}{{x}={a}+\left({a}+{d}\right)+\left({a}+\mathrm{2}{d}\right)+,…+\left({a}+\left({m}−\mathrm{1}\right){d}\right)}\\{{y}=\left({a}+{md}\right)+\left({a}+\left({m}+\mathrm{1}\right){d}\right)+…+\left({a}+\left(\mathrm{2}{m}−\mathrm{1}\right){d}\right)}\\{{z}=\left({a}+\mathrm{2}{md}\right)+\left({a}+\left(\mathrm{2}{m}+\mathrm{1}\right){d}\right)+…+\left({a}+\left(\mathrm{3}{m}−\mathrm{1}\right){d}\right)}\end{cases}\: \\ $$$$\begin{cases}{{x}−{ma}=\left({d}\right)+\left(\mathrm{2}{d}\right)+,…+\left(\left({m}−\mathrm{1}\right){d}\right)}\\{{y}−{ma}=\left({md}\right)+\left(\left({m}+\mathrm{1}\right){d}\right)+…+\left(\left(\mathrm{2}{m}−\mathrm{1}\right){d}\right)}\\{{z}−{ma}=\left(\mathrm{2}{md}\right)+\left(\left(\mathrm{2}{m}+\mathrm{1}\right){d}\right)+…+\left(\left(\mathrm{3}{m}−\mathrm{1}\right){d}\right)}\end{cases}\:\: \\ $$$$\begin{cases}{\frac{{x}−{ma}}{{d}}=\left(\mathrm{1}\right)+\left(\mathrm{2}\right)+,…+\left(\left({m}−\mathrm{1}\right)\right)}\\{\frac{{y}−{ma}}{{d}}=\left({m}\right)+\left(\left({m}+\mathrm{1}\right)\right)+…+\left(\left(\mathrm{2}{m}−\mathrm{1}\right)\right)\:}\\{\frac{{z}−{ma}}{{d}}=\left(\mathrm{2}{m}\right)+\left(\left(\mathrm{2}{m}+\mathrm{1}\right)\right)+…+\left(\left(\mathrm{3}{m}−\mathrm{1}\right)\right)}\end{cases}\: \\ $$$$…\:\: \\ $$
Answered by mr W last updated on 14/Aug/24
say the difference from the (m+1)^(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_1 , a_2 , ..., a_m ← the first m terms, their sum is x b_1 , b_2 , ..., b_m ← the next m terms, their sum is y c_1 , c_2 , ..., c_m ← the last m terms, their sum is z b_1 =a_1 +t, b_2 =a_2 +t, ..., b_m =a_m +t c_1 =b_1 +t, c_2 =b_2 +t, ..., c_m =b_m +t y=x+mt z=y+mt ⇒x+z=2y ⇒x^2 +z^2 +2xz=4y^2 ⇒x^2 +z^2 −2xz=4y^2 −4xz ⇒(x−z)^2 =4(y^2 −xz) ✓
$${say}\:{the}\:{difference}\:{from}\:{the}\:\left({m}+\mathrm{1}\right)^{{th}} \\ $$$${and}\:{the}\:{first}\:{term}\:{is}\:{t},\:{then} \\ $$$${each}\:{term}\:{in}\:{the}\:{second}\:{group}\:{is} \\ $$$${by}\:{t}\:{larger}\:{than}\:{the}\:{corresponding} \\ $$$${term}\:{in}\:{the}\:{first}\:{group}.\:{the}\:{sum}\:{of} \\ $$$${the}\:{second}\:{group}\:{is}\:{by}\:{mt}\:{larger}\:{than} \\ $$$${the}\:{sum}\:{of}\:{first}\:{group},\:{etc}. \\ $$$${a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:…,\:{a}_{{m}} \:\:\leftarrow\:{the}\:{first}\:{m}\:{terms},\:{their}\:{sum}\:{is}\:{x} \\ $$$${b}_{\mathrm{1}} ,\:{b}_{\mathrm{2}} ,\:…,\:{b}_{{m}} \:\:\leftarrow\:{the}\:{next}\:{m}\:{terms},\:{their}\:{sum}\:{is}\:{y} \\ $$$${c}_{\mathrm{1}} ,\:{c}_{\mathrm{2}} ,\:…,\:{c}_{{m}} \:\:\leftarrow\:{the}\:{last}\:{m}\:{terms},\:{their}\:{sum}\:{is}\:{z} \\ $$$${b}_{\mathrm{1}} ={a}_{\mathrm{1}} +{t},\:{b}_{\mathrm{2}} ={a}_{\mathrm{2}} +{t},\:…,\:{b}_{{m}} ={a}_{{m}} +{t} \\ $$$${c}_{\mathrm{1}} ={b}_{\mathrm{1}} +{t},\:{c}_{\mathrm{2}} ={b}_{\mathrm{2}} +{t},\:…,\:{c}_{{m}} ={b}_{{m}} +{t} \\ $$$${y}={x}+{mt} \\ $$$${z}={y}+{mt} \\ $$$$\Rightarrow{x}+{z}=\mathrm{2}{y}\: \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}{xz}=\mathrm{4}{y}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{xz}=\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{xz} \\ $$$$\Rightarrow\left({x}−{z}\right)^{\mathrm{2}} =\mathrm{4}\left({y}^{\mathrm{2}} −{xz}\right)\:\checkmark \\ $$
Commented by mm1342 last updated on 14/Aug/24
very excellent ⋛
$${very}\:\:{excellent}\:\:\:\:\:\underline{\underbrace{\lesseqgtr}} \\ $$

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