Question Number 18207 by potata8 last updated on 16/Jul/17
$$\mathrm{If}\:\frac{\mathrm{3}+\mathrm{5}+\mathrm{7}+…+{n}\:\mathrm{terms}}{\mathrm{5}+\mathrm{8}+\mathrm{11}+…+\mathrm{10}\:\mathrm{terms}}\:=\:\mathrm{7},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:{n}\:\mathrm{is} \\ $$
Answered by Tinkutara last updated on 17/Jul/17
![3 + 5 + 7 + ... + (n terms) = (n/2)[6 + (n − 1)2] = (n/2)(2n + 4) = n(n + 2) 5 + 8 + 11 + ... + (10 terms) = ((10)/2)[10 + 9×3] = 5×37 = 185 ∴ ((n(n + 2))/(185)) = 7 ⇒ n^2 + 2n − 1295 = 0 n = ((−2 ± 72)/2) = −1 ± 36 = 35 or −37 ∴ n = 35.](https://www.tinkutara.com/question/Q18218.png)
$$\mathrm{3}\:+\:\mathrm{5}\:+\:\mathrm{7}\:+\:…\:+\:\left({n}\:\mathrm{terms}\right)\:=\:\frac{{n}}{\mathrm{2}}\left[\mathrm{6}\:+\:\left({n}\:−\:\mathrm{1}\right)\mathrm{2}\right] \\ $$$$=\:\frac{{n}}{\mathrm{2}}\left(\mathrm{2}{n}\:+\:\mathrm{4}\right)\:=\:{n}\left({n}\:+\:\mathrm{2}\right) \\ $$$$\mathrm{5}\:+\:\mathrm{8}\:+\:\mathrm{11}\:+\:…\:+\:\left(\mathrm{10}\:\mathrm{terms}\right)\:= \\ $$$$\frac{\mathrm{10}}{\mathrm{2}}\left[\mathrm{10}\:+\:\mathrm{9}×\mathrm{3}\right]\:=\:\mathrm{5}×\mathrm{37}\:=\:\mathrm{185} \\ $$$$\therefore\:\frac{{n}\left({n}\:+\:\mathrm{2}\right)}{\mathrm{185}}\:=\:\mathrm{7} \\ $$$$\Rightarrow\:{n}^{\mathrm{2}} \:+\:\mathrm{2}{n}\:−\:\mathrm{1295}\:=\:\mathrm{0} \\ $$$${n}\:=\:\frac{−\mathrm{2}\:\pm\:\mathrm{72}}{\mathrm{2}}\:=\:−\mathrm{1}\:\pm\:\mathrm{36}\:=\:\mathrm{35}\:\mathrm{or}\:−\mathrm{37} \\ $$$$\therefore\:\boldsymbol{{n}}\:=\:\mathrm{35}. \\ $$