Applied problems in sequence of numbers Since 1986, the total population and the total residential area of a certain area have been increasing year by year according to the equal ratio series and the equal difference series respectively. It is known that the per capita housing area at the end of 1986 is 10m ^ 22006, and the per capita housing area at the end of 1986 is 20m ^ 2. According to this calculation, the per capita housing area at the end of 1996 is more than 14m ^ 2, and the proof is given; 2, Is it ensured that the per capita housing area at the end of 2008 will increase compared with that at the end of 2006? Why?

Applied problems in sequence of numbers Since 1986, the total population and the total residential area of a certain area have been increasing year by year according to the equal ratio series and the equal difference series respectively. It is known that the per capita housing area at the end of 1986 is 10m ^ 22006, and the per capita housing area at the end of 1986 is 20m ^ 2. According to this calculation, the per capita housing area at the end of 1996 is more than 14m ^ 2, and the proof is given; 2, Is it ensured that the per capita housing area at the end of 2008 will increase compared with that at the end of 2006? Why?

Let the total population of 1986 be A1 and that of 2006 be A21
Suppose that the total residential area in 1986 is B1 = 10 * A1, and that in 2006 is B21 = 20 * A21
Then A1 * q ^ 20 = A21, B1 + 20 * d = B21, then q = (A21 / A1) to the power of 20, d = a21-1 / 2 * A1
1. Prove: B11 = B1 + 10 * d = 5 * a1 + 10 * A21, a11 = A1 * q ^ 10 = open radical (A1 * A21)
B11 > = 2 * radical (5 * A1 * 10 * A21) = 10 * radical (2 * A1 * A21) > = 14 * radical (A1 * A21)
Therefore, at the end of 1996, the per capita housing area exceeded 14 m ^ 2
2. The average annual growth rate of population is no more than 3%, i.e. Q