Question

Solve the question $\left(-4\right)+\left(-1\right)+2+5+....+x=437.$

Answer 1

Suppose x is nth term of the given A.P.
$$\begin{gathered} a_n=x \\ Here,a=-4,d=3. \\ Itisgiventhat,S_n=437. \\ \Rightarrow \frac{n}{2}\left[2\left(-4\right)+\left(n-1\right)3\right]=437 \\ \Rightarrow 3n^2-11n-874=0 \\ \Rightarrow 3n^2-57n+46n-874=0 \\ \Rightarrow 3n\left(n-19\right)+46\left(n-19\right)=0 \\ \Rightarrow n=-\frac{46}{3},19 \\ Since,ncannotbeinfractionson=19. \\ Now,a_n=x \\ \Rightarrow \left(-4\right)+\left(19-1\right)3=x \\ \Rightarrow -4+54=x \\ \Rightarrow x=50 \end{gathered}$$