imc

Let $x=a+1=b+2=c+3=d+4=a+b+c+d+5$. Since one of the sums involves $a,b,c,$ and $d,$ it makes sense to consider $4x$. We have $4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20$. Rearranging, we have $3(a+b+c+d)=-10$, so $a+b+c+d=\frac{-10}{3}$. Thus, our answer is $\boxed{-10/3}$.