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Question Number 18207 by potata8 last updated on 16/Jul/17

$$\mathrm{If}\frac{3+5+7+...+n\mathrm{terms}}{5+8+11+...+10\mathrm{terms}}=7,\mathrm{then}\mathrm{the}$$$$\mathrm{value}\mathrm{of}n\mathrm{is}$$

Answered by Tinkutara last updated on 17/Jul/17

$$3+5+7+...+\left(n\mathrm{terms}\right)=\frac{n}{2}[6+(n-1)2]$$$$=\frac{n}{2}(2n+4)=n(n+2)$$$$5+8+11+...+\left(10\mathrm{terms}\right)=$$$$\frac{10}{2}[10+9\times 3]=5\times 37=185$$$$\therefore \frac{n(n+2)}{185}=7$$$$\Rightarrow n^2+2n-1295=0$$$$n=\frac{-2\pm 72}{2}=-1\pm 36=35\mathrm{or}-37$$$$\therefore \mathit{n}=35.$$

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