Question

如果那么 $N =$ $\begin{aligned} &\frac{2+3+4}{3} \\ &= \frac{1990+1991+1992}{N}, \end{aligned}$

If $\begin{aligned} &\frac{2+3+4}{3} \\ &= \frac{1990+1991+1992}{N}, \end{aligned}$ then $N =$

$3$

$6$

$1990$

$1991$

$1992$

左边是 $\dfrac{9}{3} = 3$ 。右边是 $\dfrac{5973}{N}$ ，令它等于 $3$ ，得到 $N = 1991$ 。

也可以用规律： $(k-1) + k + (k+1) = 3k$ ，所以除以 $3$ 后得到中间项。这里中间项是 $1991$ 。

所以正确答案是 D。

The left side is $\dfrac{9}{3} = 3.$ The right side is $\dfrac{5973}{N},$ and setting it equal to $3$ gives $N = 1991.$

Equivalently, $(k-1) + k + (k+1) = 3k,$ so dividing by $3$ leaves the middle term. Here the middle term is $1991.$

Thus, the correct answer is D .