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Question Number 140834 by greg_ed last updated on 13/May/21
hi, sirs !  please, how to solve this thing below (a ∈ R)            { ((x_1 −x_2  = a)),((x_2 −x_3  = 2a)),((.................)),((.................)),((................)),((x_(n−1) − x_n  = (n−1)a)),((x_1 + x_2 + ... + x_(n−1) + x_n  = na)) :}
$$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{sirs}}\:! \\ $$$$\boldsymbol{\mathrm{please}},\:\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{thing}}\:\boldsymbol{\mathrm{below}}\:\left(\boldsymbol{{a}}\:\in\:\mathbb{R}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\begin{cases}{\boldsymbol{{x}}_{\mathrm{1}} −\boldsymbol{{x}}_{\mathrm{2}} \:=\:\boldsymbol{{a}}}\\{\boldsymbol{{x}}_{\mathrm{2}} −\boldsymbol{{x}}_{\mathrm{3}} \:=\:\mathrm{2}\boldsymbol{{a}}}\\{……………..}\\{……………..}\\{…………….}\\{\boldsymbol{{x}}_{\boldsymbol{{n}}−\mathrm{1}} −\:\boldsymbol{{x}}_{\boldsymbol{{n}}} \:=\:\left(\boldsymbol{{n}}−\mathrm{1}\right)\boldsymbol{{a}}}\\{\boldsymbol{{x}}_{\mathrm{1}} +\:\boldsymbol{{x}}_{\mathrm{2}} +\:…\:+\:\boldsymbol{{x}}_{\boldsymbol{{n}}−\mathrm{1}} +\:\boldsymbol{{x}}_{\boldsymbol{{n}}} \:=\:\boldsymbol{{na}}}\end{cases} \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 13/May/21
Adding all the equations:  x_1 +x_n =a(1+2+...+n)  x_1 +x_n =((n(n+1)a)/2)  .....  ...
$${Adding}\:{all}\:{the}\:{equations}: \\ $$$${x}_{\mathrm{1}} +{x}_{{n}} ={a}\left(\mathrm{1}+\mathrm{2}+…+{n}\right) \\ $$$${x}_{\mathrm{1}} +{x}_{{n}} =\frac{{n}\left({n}+\mathrm{1}\right){a}}{\mathrm{2}} \\ $$$$….. \\ $$$$… \\ $$$$ \\ $$
Commented by prakash jain last updated on 13/May/21
  x_1 =x_1   x_2 =x_1 −a  x_3 =x_2 −2a=x_1 −(a+2a)  x_4 =x_1 −(a+2a+3a)  ...  x_n =x_1 −(((n−1)n)/2)a  Σ_(i=1) ^n x_i =nx_1 −Σ_(i=1) ^n i(i−1)a  =nx_1 −aΣ_(i=1) ^n i^2 −aΣ_(i=1) ^n i  =nx_1 −((n(n+1)(2n+1)a)/6)−((n(n+1))/2)a  =na (given)  x_1 =a+(((n+1))/2)a+(((n+1)(2n+1))/6)a  =(a/3)(n^2 +3n+5)  x_i =x_1 −((i(i−1))/2)×a
$$ \\ $$$${x}_{\mathrm{1}} ={x}_{\mathrm{1}} \\ $$$${x}_{\mathrm{2}} ={x}_{\mathrm{1}} −{a} \\ $$$${x}_{\mathrm{3}} ={x}_{\mathrm{2}} −\mathrm{2}{a}={x}_{\mathrm{1}} −\left({a}+\mathrm{2}{a}\right) \\ $$$${x}_{\mathrm{4}} ={x}_{\mathrm{1}} −\left({a}+\mathrm{2}{a}+\mathrm{3}{a}\right) \\ $$$$… \\ $$$${x}_{{n}} ={x}_{\mathrm{1}} −\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}{a} \\ $$$$\sum_{{i}=\mathrm{1}} ^{{n}} {x}_{{i}} ={nx}_{\mathrm{1}} −\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}\left({i}−\mathrm{1}\right){a} \\ $$$$={nx}_{\mathrm{1}} −{a}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}^{\mathrm{2}} −{a}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i} \\ $$$$={nx}_{\mathrm{1}} −\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right){a}}{\mathrm{6}}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}{a} \\ $$$$={na}\:\left({given}\right) \\ $$$${x}_{\mathrm{1}} ={a}+\frac{\left({n}+\mathrm{1}\right)}{\mathrm{2}}{a}+\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}{a} \\ $$$$=\frac{{a}}{\mathrm{3}}\left({n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{5}\right) \\ $$$${x}_{{i}} ={x}_{\mathrm{1}} −\frac{{i}\left({i}−\mathrm{1}\right)}{\mathrm{2}}×{a} \\ $$
Commented by Rasheed.Sindhi last updated on 13/May/21
W∈LCOM∈ SIR  after long interval!
$$\mathcal{W}\in\mathcal{LCOM}\in\:\mathcal{SIR}\:\:{after}\:{long}\:{interval}! \\ $$
Commented by Ar Brandon last updated on 13/May/21
Greetings Sir Prakash Jain.  😃
$$\mathrm{Greetings}\:\mathrm{Sir}\:\mathrm{Prakash}\:\mathrm{Jain}. \\ $$😃

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