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Question Number 140834 by greg_ed last updated on 13/May/21

$$\mathbf{hi},\mathbf{sirs}!$$$$\mathbf{please},\mathbf{how}\mathbf{to}\mathbf{solve}\mathbf{this}\mathbf{thing}\mathbf{below}(\mathit{a}\in \mathbb{R})$$$$\{\begin{array}{l}{\mathit{x}}_1-{\mathit{x}}_2=\mathit{a}\\ {\mathit{x}}_2-{\mathit{x}}_3=2\mathit{a}\\ .................\\ .................\\ ................\\ {\mathit{x}}_{\mathit{n}-1}-{\mathit{x}}_{\mathit{n}}=(\mathit{n}-1)\mathit{a}\\ {\mathit{x}}_1+{\mathit{x}}_2+...+{\mathit{x}}_{\mathit{n}-1}+{\mathit{x}}_{\mathit{n}}=\mathit{n}\mathit{a}\end{array}$$$$$$$$$$

Commented by Rasheed.Sindhi last updated on 13/May/21

$$Addingalltheequations:$$$$x_1+x_n=a(1+2+...+n)$$$$x_1+x_n=\frac{n(n+1)a}{2}$$$$.....$$$$...$$$$$$

Commented by prakash jain last updated on 13/May/21

$$$$$$x_1=x_1$$$$x_2=x_1-a$$$$x_3=x_2-2a=x_1-(a+2a)$$$$x_4=x_1-(a+2a+3a)$$$$...$$$$x_n=x_1-\frac{(n-1)n}{2}a$$$$\sum _{i=1}^n{x}_i={nx}_1-\underset{i=1}{\sum ^n}i(i-1)a$$$$={nx}_1-a\underset{i=1}{\sum ^n}{i}^2-a\underset{i=1}{\sum ^n}i$$$$={nx}_1-\frac{n(n+1)(2n+1)a}{6}-\frac{n(n+1)}{2}a$$$$=na\left(given\right)$$$$x_1=a+\frac{(n+1)}{2}a+\frac{(n+1)(2n+1)}{6}a$$$$=\frac{a}{3}(n^2+3n+5)$$$$x_i=x_1-\frac{i(i-1)}{2}\times a$$

Commented by Rasheed.Sindhi last updated on 13/May/21

$$\mathcal{W}\in \mathcal{LCOM}\in \mathcal{SIR}afterlonginterval!$$

Commented by Ar Brandon last updated on 13/May/21

$$\mathrm{Greetings}\mathrm{Sir}\mathrm{Prakash}\mathrm{Jain}.$$😃

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