Solve the equation

Given equation is, – 4 -1 + 2 + … + x = 437                                                                                  …(i)

Here, – 4-1 + 2 + …+ x forms an AP with first term =-4, common difference =3,

an = l = x

$\because$ nth term of an AP, $a_{n}=l=a+(n-1) d$

$\Rightarrow \quad x=-4+(n-1) 3$

$\Rightarrow \quad \frac{x+4}{3}=n-1 \Rightarrow n=\frac{x+7}{3}$ ....(ii)

$\therefore \quad$ Sum of an AP, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{n}=\frac{x+7}{2 \times 3}\left[2(-4)+\left(\frac{x+4}{3}\right) \cdot 3\right]$

$=\frac{x+7}{2 \times 3}(-8+x+4)=\frac{(x+7)(x-4)}{2 \times 3}$

From Eq. (i),

$S_{n}=437$

$\Rightarrow \quad \frac{(x+7)(x-4)}{2 \times 3}=437$

$\Rightarrow \quad x^{2}+7 x-4 x-28=874 \times 3$

$\Rightarrow \quad x^{2}+3 x-2650=0$

$x=\frac{-3 \pm \sqrt{(3)^{2}-4(-2650)}}{2}$ [by quadratic formula]

$=\frac{-3 \pm \sqrt{9+10600}}{2}$

Here, x cannot be negative, i.e., x ≠- 53

also, for x = -53, n will be negatuve which is not possible

Hence, the required value of x is 50.