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Question Number 78979 by jagoll last updated on 22/Jan/20

$$\mathrm{the}\mathrm{first}\mathrm{n}\mathrm{even}$$$$\mathrm{numbers}\mathrm{are}\mathrm{known}.\mathrm{if}\mathrm{one}\mathrm{number}$$$$\mathrm{is}\mathrm{deleted}\mathrm{then}\mathrm{the}\mathrm{average}$$$$\mathrm{of}\mathrm{remaining}\mathrm{numbers}\mathrm{is}\frac{2582}{50}.$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{deleted}\mathrm{number}.$$

Commented by mr W last updated on 22/Jan/20

$$the51numbersare2,4,6,...102$$$$thedeletetenumberis70$$

Commented by jagoll last updated on 22/Jan/20

$$\mathrm{sir}\mathrm{please}\mathrm{explain}\mathrm{me}$$

Commented by mr W last updated on 22/Jan/20

$$nevennumbers:2(1+2+...+n)=n(n+1)$$$$deletednumber:2mwithm⩽n$$$$\frac{n(n+1)-2m}{n-1}=\frac{2582}{50}=\frac{1291k}{25k}$$$$n(n+1)-\frac{1291}{25}(n-1)=2m⩽2n$$$$n^2-\frac{1316}{25}n+\frac{1291}{25}⩽0$$$$\Rightarrow 1⩽n⩽51$$$$\Rightarrow 1⩽k⩽2$$$$k=1:\Rightarrow n=26\Rightarrow nointegerm$$$$k=2:\Rightarrow n=51\Rightarrow m=35\Rightarrow solution$$

Commented by jagoll last updated on 22/Jan/20

$$\mathrm{thanks}\mathrm{you}\mathrm{lot}\mathrm{sir}$$

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