2002 AMC 10A Problem 16

Let the common value be $k.$ Then $a=k-1,$ $b=k-2,$ $c=k-3,$ $d=k-4,$ so $a+b+c+d=4k-10.$ Since $a+b+c+d+5=k,$ we get $4k-10+5=k,$ so $3k=5$ and $k=\dfrac{5}{3}.$ Then $a+b+c+d=k-5$ $=\dfrac{5}{3}-5=-\dfrac{10}{3}.$ Thus, the correct answer is B.