Senior three math problems, senior one can also be counted, senior three math problems?

27, I have a method that doesn't need to calculate a1:

Because a3+a4=9, a1+a2=1, the common ratio is 3 (a3+a4=a1*q 2+a2*q 2) (q 2 is the square of q).

After getting the common ratio, there is a4+a5=a3*q+a4*q=9*3=27, I admit that I did it wrong, and q=-3 is also true, so a4+a5=27 or -27 will be good to synthesize my previous practice with cooking millet in a pot.

Hey, I'm confused again, the answer is indeed 27, because the question says an>0, so q=-3 is wrong. The answer is 27

Start by listing the equations with the first few known conditions to find the common ratio = 3 The first term is a quarter. Then you can calculate that the answer is 27

a1 + a2, a3 + a4, a4 + a5 into a new proportional series, the first term a1 + a2, the common ratio is 9,,, so a4 + a5 is 27

Agree with yuguolege's approach, it's more ingenious.

Let the general formula for the difference series be an = a1 + n-1)d, where d is the tolerance and a1 is the first term.

From the title can be known to the people:

a2, a5, a7 form a proportional sequence, ie.

a5 / a2 = a7 / a5

Substituting the general formula of the difference series yields:

a1 + 4d) /a1 + d) =a1 + 6d) /a1 + 4d)

The solution gives d = 1 29.

Substituting 2a1 + a2 = 1 and a2 = a1 + d yields:

2a1 + a1 + d = 1

3a1 = 1 - d

The solution gives a1 = 10 29.

Therefore, the general formula for a series of equal differences is:

an = 10 29 - n-1) liter 29 so, a10 = 10 29 - 9 29 = 1 29.

Summing the A1 and A10 generations of hood reeds into the equation, we get:

s10 = 10/29 + 1/29) ×10/2 = 11/29 × 5 = 55/29

Therefore, s10 is equal to 55 29.

1) The idea can be that the cosine theorem turns the angular into a side finishing, without trying.

Here the sine is used to definite the answer.

2) Using the relationship + mean inequality of a, b, c in (1), the process is as follows.

You can use the special value method, first bring the coordinates of point A into the elliptic equation to know that A is on the ellipse, and then you can set Kaq and KAP as 1 and -1 respectively, list the equations of AQ and AP, substitute the coordinates of point A to get two straight line equations, and the simultaneous linear equation and elliptic equation can obtain the coordinates of the intersection point, one of the intersection points is A, take the other two intersection points, that is, the coordinates of PQ, let the straight line PQ equation: y=kx+b, and substitute the two coordinates to find the PQ equation, where K is the slope. Pure hand-played, non-copying.

For simple fill-in-the-blank questions, it is recommended to use the special value method. Because the system approach is more complex, it is as follows: To be continued.

Tell you, if two straight lines with complementary angles of inclination intersect at two points b and c at any point of the ellipse, a(x0,y0), then the slope k of the straight line is (b x0) (a y0), if you need to derive it, you will return to me.

<> saw it, I hope it can help you.

I'm in the fourth grade, and I don't have time, so I don't want to go on my own.

Because it's not good at typing, it's written directly by hand.

The handwriting is not good, please forgive me (I really tried my best).

If I'm not mistaken, this is it, you can refer to the process.

The first question is the median line nature of the triangle, and the second question uses the equal volume method.