Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

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))

2S=n[2a+nd−d]⇒2S=2an+n2d−dn⇒n2d+(2a−d)n−2S=0n=(d−2a)±(2a−d)2+8dS2d

Terms of Service

Privacy Policy

Contact: info@tinkutara.com