Question

solve -4+[-1]+2........+x=437

Answer 1

It is an arithmetic series.

a = -4    d = +3
-4  - 1  + 2  + 5 + .... + (x-3) + x = 437

Sum of the series = [ 2 a + (n-1)d ]n /2 = 437 
                        [ - 8 + 3(n-1) ] n / 2  = 437
                   - 11n + 3n² = 874
           3 n² -11n - 874 = 0
               n =   (11 +- 103 )/6  =  114/6 = 19   as n is +ve only.

So x = a + (n-1)d = -4 + 3 (19-1) = 50

Answer 2

first term(a) = - 4    
common difference(d) = 3
- 4  - 1 + 2 + 5 + .... + x = 437

so now using formula for sum of n terms
$S_{n}=\frac{n}{2}[2a+(n-1)d]$

⇒437= n/2[ -8 + (n-1)3]
⇒n[ -8 + 3(n-1) ] = 874
⇒ -11x=n + 3n² = 874
⇒3n² -11n - 874 = 0
Using sidharacharya
no. of terms cannot be negative the we take the positive value
n = 19
So
 x = a + (n-1)d = -4 + 3 (19-1) = 50