2.a 1+(n- 1)d=33

1/3+(n- 1)d=33

Also, A2+A5 = 4, A 1+D+A 1+4D = 4.

Solution, d=2/3. Substitution in the first formula gives n=50.

3. Items in application, a 1+a2+a3= 15, that is, 2a2+a2= 15, a2=5.

A 1a2a3=80, so a 1a3= 16, simultaneous a 1+a3= 10, the calculation result is a 1=2.a3=8 (the tolerance is positive.

D=3，a 1 1+a 12+a 13 = a 1+ 10d+A2+ 10d+A3+ 10d = 65438+。

4.2b=a+c (arithmetic mean)

The discriminant of the equation is 2b squared -4ac,

Substituting the above formula, the complete square of (a-c) is simplified. It is not negative, that is, it is greater than or equal to 0, so there are 1 or 2 roots.