Question Number 172999 by Mikenice last updated on 04/Jul/22
Answered by MikeH last updated on 05/Jul/22
![2S = n[2a + nd−d] ⇒ 2S = 2an + n^2 d − dn ⇒ n^2 d + (2a−d)n−2S = 0 n = (((d−2a)±(√((2a−d)^2 +8dS)))/(2d))](https://www.tinkutara.com/question/Q173034.png)
$$\mathrm{2}{S}\:=\:{n}\left[\mathrm{2}{a}\:+\:{nd}−{d}\right] \\ $$$$\Rightarrow\:\mathrm{2}{S}\:=\:\mathrm{2}{an}\:+\:{n}^{\mathrm{2}} {d}\:−\:{dn} \\ $$$$\Rightarrow\:{n}^{\mathrm{2}} {d}\:+\:\left(\mathrm{2}{a}−{d}\right){n}−\mathrm{2}{S}\:=\:\mathrm{0} \\ $$$${n}\:=\:\frac{\left({d}−\mathrm{2}{a}\right)\pm\sqrt{\left(\mathrm{2}{a}−{d}\right)^{\mathrm{2}} +\mathrm{8}{dS}}}{\mathrm{2}{d}} \\ $$