Question

Solve the equation : $-4+\left(-1\right)+2+\cdot \cdot \cdot \cdot x=437.$

Answer 1

Given the series $[-4+(-1)+2+\cdot \cdot \cdot \cdot x=437]$ in A.P.

Then,

$a=-4$

$d=-1-(-4)=3$

$S_n=437$

Then, using sum formula A.P.

$S_n=\frac{n}{2}[2a+(n-1)d]$

$\Rightarrow 437=\frac{n}{2}[2(-4)+(n-1)3]$

$\Rightarrow 874=n[-8+3n-3]$

$\Rightarrow 874=3n^2-11n$

$\Rightarrow 3n^2-11n=874$

$\Rightarrow 3n^2-11n-874=0$

$\Rightarrow 3n^2-(57-46)n-874=0$

$\Rightarrow 3n^2-57n-46n-874=0$

$\Rightarrow 3n(n-19)-46(n-19)=0$

$\Rightarrow (n-19)(3n-19)=0$

$\Rightarrow n-19=0,3n-19=0$

$\Rightarrow n=19,n=\frac{19}{3}$

Hence, $n=19$

So,

$x=a+(n-1)d$

$x=-4+(19-1)3$

$x=-4+54$

$x=50$

Hence, this is the answer.