Paul has to replace 3000 m of fencing on his farm.

Let `F_n` be the length, in metres, of fencing left to replace after `n` weeks.

The difference equation

`F_(n + 1) = 0.95F_n + a\ \ \ \ \ \ F_0 = 3000`

can be used to calculate the length of fencing left to replace after `n` weeks.

In this equation, `a` is a constant.

After one week, Paul still has 2540 m of fencing left to replace.

After three weeks, the length of fencing, in metres, left to replace will be closest to

**A.** 1310

**B. ** 1380

**C.** 1620

**D.** 1690

**E. ** 2100