Suppose that \(M\) is a Abelian group in which the unique division by 2 and 3 is performable and \(S\) is an abstract cone satisfying the cancellation law. In this paper the author proves that if \(f:M\to S\) is a solution of the Jensen functional equation, then it is a solution of the following equation \[ 3(b-1) f\biggl(\frac{x+y+z}{3}\biggr)+ f(x)+f(y)+f(z)= b\Biggl[f\biggl(\frac{x+y}{2}\biggr)+ f\biggl(\frac{y+z}{2}\biggr)+ f\biggl(\frac{z+x}{2}\biggr)\Biggr] \] and conversely, if \(b\geq 1, b\neq 4\) and \(f:M\to S\) is a solution of the equation above, then \(f\) is a solution of the Jensen functional equation. This result is based on works of \textit{Y. W. Lee} [J. Math. Anal. Appl. 270, No. 2, 590--601(2002; Zbl 1007.39026)] and \textit{T. Trif} [J. Math. Anal. Appl. 250, No. 2, 579--588 (2000; Zbl 0964.39027)]. He also shows that the function \(f\) satisfies the equation above for \(b=4\) if and only if it fullfills the system of functional equations \[ \begin{aligned} f(x+y)+f(-x)+f(-y)&= f(-x-y)+f(x)+f(y),\\ 2f(0)+f(x+y)+f(x-y)&= 2f(x)+f(y)+f(-y).\end{aligned} \]