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Question Number 5125 by Rojaye Shegz last updated on 16/Apr/16

Commented by Rojaye Shegz last updated on 16/Apr/16

$$\mathrm{Sorry},\mathrm{that}\mathrm{was}\mathrm{an}\mathrm{error}.$$$$\frac{\mathrm{p}}{\mathrm{q}}\mathrm{is}\mathrm{the}\{\frac{1}{2}(\mathrm{p}+\mathrm{q}-1)(\mathrm{p}+\mathrm{q}-2)+\mathrm{p}\}\mathrm{th}\mathrm{term}$$

Commented by 123456 last updated on 17/Apr/16

$$\frac{1}{1}\Rightarrow 1+1=2\left(1\right)$$$$\frac{1}{2},\frac{2}{1}\Rightarrow 1+2=3\left(2\right)$$$$\frac{1}{3},\frac{2}{2},\frac{3}{1}\Rightarrow 1+3=4\left(3\right)$$$$\frac{1}{4},\frac{2}{3},\frac{3}{2},\frac{4}{1}\Rightarrow 1+4=5\left(4\right)$$$$⋮$$$$\frac{1}{q},\frac{2}{q-1},\frac{3}{q-2},\dots ,\frac{q-2}{3},\frac{q-1}{2},\frac{q}{1}\Rightarrow p+q=k(p+q-1)$$$$\underset{i=1}{\sum ^{k-1}}(i-1)=\frac{(k-1-1+1)(k-1-1+1-1)}{2}$$$$=\frac{(k-1)(k-2)}{2}$$$$=\frac{(a+b-1)(a+b-2)}{2}$$$$\frac{1}{q},\frac{2}{q-1},...,\frac{p}{q},...,\frac{q-1}{2},\frac{q}{1}\Rightarrow p$$$$\frac{(a+b-1)(a+b-2)}{2}+p$$

Commented by Rojaye Shegz last updated on 17/Apr/16

$$\mathrm{I}\mathrm{dont}\mathrm{understand}\mathrm{your}\mathrm{solution}$$$$\mathrm{but}\mathrm{im}\mathrm{gaining}\mathrm{insight}\mathrm{on}\mathrm{how}\mathrm{it}\mathrm{should}$$$$\mathrm{be}\mathrm{solved}.\mathrm{Please}\mathrm{explain}$$$${\mathrm{Thanks}}^{+}$$

Answered by Yozzii last updated on 16/Apr/16

$$For{u}_3=\frac{2}{1}=\frac{p}{q}$$$$\Rightarrow q=1\mathrm{\&}p=2.$$$$\therefore 0.5(3-1)(3-2)+1=0.5\times 2+1=1+1=2\ne 3$$$$For{u}_2=\frac{1}{2}\Rightarrow p=1,q=2$$$$\therefore 0.5(3-1)(3-2)+2=3\ne 2.$$

Answered by 123456 last updated on 17/Apr/16

$$\mathrm{lets}\mathrm{organize}\mathrm{the}\mathrm{sequence}\mathrm{on}\mathrm{this}\mathrm{way}$$$$\frac{1}{1}$$$$\frac{1}{2},\frac{2}{1}$$$$\frac{1}{3},\frac{2}{2},\frac{3}{1}$$$$\frac{1}{4},\frac{2}{3},\frac{3}{2},\frac{4}{1}$$$$⋮$$$$\mathrm{each}\mathrm{colum}\mathrm{have}a+b=\mathrm{comstant}\mathrm{with}$$$$\frac{a}{b},\mathrm{and}\mathrm{its}\mathrm{is}i+1\left(\mathrm{number}\mathrm{of}\mathrm{colum}\mathrm{plus}\mathrm{one}\right)$$$$\mathrm{lets}\mathrm{have}\frac{p}{q},\mathrm{conting}\mathrm{all}\mathrm{elements}\mathrm{before}$$$$\mathrm{this}\mathrm{colum}$$$$\mathrm{the}\mathrm{colum}\mathrm{of}p/q\mathrm{is}p+q-1,\mathrm{so}\mathrm{one}\mathrm{before}\mathrm{is}$$$$p+q-2$$$$\underset{i=1}{\sum ^{p+q-2}}i=\frac{(p+q-2)(p+q-2+1)}{2}(P.A)$$$$=\frac{(p+q-2)(p+q-1)}{2}$$$$\mathrm{now}\mathrm{lets}\mathrm{count}\mathrm{element}\mathrm{at}\mathrm{colum}p+q-1$$$$\frac{1}{q},\frac{2}{q-1},...,\frac{p}{q}$$$$\mathrm{as}\mathrm{you}\mathrm{can}\mathrm{see},\mathrm{at}p/qwehavep\mathrm{elements}$$$$\mathrm{so}\mathrm{the}\mathrm{total}\mathrm{of}\mathrm{elements}\mathrm{is}\left(\mathrm{including}\mathrm{it}\right)$$$$\frac{(p+q-2)(p+q-1)}{2}+p$$$$\mathrm{wich}\mathrm{make}\mathrm{it}$$$$\frac{(p+q-2)(p+q-1)}{2}+p\mathrm{nth}\mathrm{element}$$

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