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Question Number 34734 by mondodotto@gmail.com last updated on 10/May/18

$$\mathbf{given}\mathbf{the}\mathbf{sum}\mathbf{of}\mathbf{the}\mathbf{first}\mathbf{n}\mathbf{terms}$$$$\mathbf{of}\mathbf{an}\mathbf{AP}\mathbf{is}{\mathit{x}}^2\mathbf{the}\mathbf{sum}\mathbf{of}$$$$\mathbf{the}\mathbf{first}2\mathbf{n}\mathbf{terms}\mathbf{of}\mathbf{the}\mathbf{same}\mathbf{AP}\mathbf{is}{\mathit{x}}^2+\mathit{x}$$$$\mathbf{show}\mathbf{that}\mathbf{the}\mathbf{sum}\mathbf{of}\mathbf{the}\mathbf{first}4\mathbf{n}\mathbf{terms}\mathrm{is}$$$$4{\mathit{x}}^2-8\mathit{x}+4$$

Commented by abdo mathsup 649 cc last updated on 11/May/18

$$ifthefirsttermis{u}_{1}wehave$$$$\sum _{k=1}^n{u}_k=x^2\Rightarrow \frac{n}{2}(u_1+u_n)=x^2\Rightarrow$$$$u_1+u_n=\frac{2x^2}{n}\Rightarrow u_1+(n-1)r=\frac{2x^2}{n}\Rightarrow$$$$(n-1)r=\frac{2x^2}{n}-u_1\Rightarrow r=\frac{2x^2-{nu}_1}{n-1}$$$$rdependsonnrisnotconstanttheQ.contain$$$$aerror....$$

Answered by Rasheed.Sindhi last updated on 10/May/18

$$a:\mathrm{first}\mathrm{term},d:\mathrm{common}\mathrm{difference}$$$$\mathrm{Sum}\mathrm{of}\mathrm{n}\mathrm{terms}:$$$$\frac{n}{2}[2a+\overline{n-1}d]=x^2$$$$2an+n(n-1)d=2x^2....\mathrm{I}$$$$\mathrm{Sum}\mathrm{of}2\mathrm{n}\mathrm{terms}:$$$$\frac{2n}{2}[2a+\overline{2n-1}d]=x^2+x$$$$2an+n(2n-1)d]=x^2+x...\mathrm{II}$$$$\mathrm{I}-\mathrm{II}:[n(n-1)-n(2n-1)]d=x^2-x$$$$n(n-1-2n+1)d=x^2-x$$$$n(-n)d=x^2-x$$$$-n^{2}d=x^2-x$$$$d=\frac{x-x^2}{n^2}$$$$\frac{n}{2}[2a+\overline{n-1}d]=x^2$$$$\Rightarrow n[2a+(n-1)\left(\frac{x-x^2}{n^2}\right)]=2x^2$$$$2an=2x^2-\frac{n-1}{n^2}(x-x^2)$$$$a=\frac{x^2}{n}-\frac{n-1}{2n^3}(x-x^2)$$$$a=\frac{2n^2{x}^2-(n-1)(x-x^2)}{2n^3}$$$$\mathrm{Sum}\mathrm{of}4\mathrm{n}\mathrm{terms}:$$$$\frac{4\mathrm{n}}{2}[2a+(4n-1)d]$$$$=2n[2\left(\frac{2n^2{x}^2-(n-1)(x-x^2)}{2n^3}\right)+(4n-1)\left(\frac{x-x^2}{n^2}\right)]$$$$=2\left(\frac{2n^2{x}^2-(n-1)(x-x^2)}{n^2}\right)+2n(4n-1)\left(\frac{x-x^2}{n^2}\right)$$$$=2\left(\frac{2n^2{x}^2-(n-1)(x-x^2)+2n(4n-1)(x-x^2)}{n^2}\right)$$$$=\frac{4n^2{x}^2-2(n-1)(x-x^2)+4n(4n-1)(x-x^2)}{n^2}$$$$=\frac{4n^2{x}^2-2nx+2{nx}^2+2x-2x^2+16n^{2}x-16n^2{x}^2-4nx+4{nx}^2}{n^2}$$$$\mathrm{The}\mathrm{answer}\mathrm{is}\mathrm{not}\mathrm{free}\mathrm{of}\mathrm{n}$$$$\mathrm{Perhsps}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{an}\mathrm{error}...$$$$\mathrm{Can}\mathrm{you}\mathrm{check}\mathrm{my}\mathrm{solution}.$$$$Continue$$

Commented by mondodotto@gmail.com last updated on 10/May/18

$$\mathrm{please}\mathrm{finish}\mathrm{it}\mathrm{up}$$

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