Question

Solve the equation : 1+4+7+10+...+x=287

Answer 1

$\begin{array}{c}Here,given\\ a=1\\ d=4-1=3\\ and,s_n=287\\ Now,\\ s_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\\ \Rightarrow 287=\frac{n}{2}\left(2\times 1+\left(n-1\right)3\right)\\ \Rightarrow 287=\frac{n}{2}\left(2+3n-3\right)\\ \Rightarrow 574=n\left(3n-1\right)\\ \Rightarrow 574=3n^2-n\\ \Rightarrow 3n^2-n-574=0\\ onsolvingthequadraticequatonu\sin gformula\\ n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ Wegetn=14\&\frac{-41}{3}\left[doesnotexist\right]\\ so,n=14\\ Now,\\ s_n=\frac{n}{2}\left(a+1\right)\\ \Rightarrow 287=\frac{14}{2}\left(1+x\right)\\ \Rightarrow 574=14(1+x)\\ \Rightarrow \left(1+x\right)=\frac{574}{14}\\ \Rightarrow 1+x=41\\ \Rightarrow x=41-1\\ \therefore x=40\\ x=40isthesolution.\end{array}$