The method of finding the term and sum in the equal ratio sequence Middle term and sum

The method of finding the term and sum in the equal ratio sequence Middle term and sum

The square of a term in an equal ratio sequence = its preceding term * its following term
Sum = A1 (1-Q ^ n) / (1-Q) = (A1 anq) / (1-Q)

How to find the middle term of equal ratio?
We're learning these days. I don't understand that big brother. Help me
A top of gratitude

Should be before and after the two top multiplication and then square

How to deduce the solution of the middle term of equal ratio

If the middle term of the ratio is am, the ratio Q
According to the definition of equal ratio sequence, it is easy to get that the former term is am / Q and the latter term is am * Q
Then am / Q * am * q = am ^ 2

Equal ratio series!
1 1-q^3
— = ————
9 1-q^6
How to simplify it like this!

1 1-q^3
— = ————
9 1-q^6
1/9=（1-q^3)/[(1-q^3)(1+q^3)]=1/(1+q^3)
1+q^3=9 q^3=8 q=2

There are two rows of seats, 11 in the front row and 12 in the back row. Now two people are seated. It is stipulated that the three seats in the middle of the front row are not allowed to sit, and the two people are not adjacent to each other, so the number of different arrangements is ()
A. 234B. 346C. 350D. 363

(1) the three seats in the middle of the front row are not allowed to sit, and the two people are not adjacent to each other, one in the front row and one in the back row, with a total of 2c81 · C121 = 192. (2) there are two seats in the back row (not adjacent), 2 (10 + 9 + 8 +...) +1) (3) two 2 (6 + 5 +) seats in the front row +1) There are 192 + 110 + 44 = 346

A7n-a5n / A5N = 89, what is the value of N

A7n=n(n-1)…… （n-4)（n-5)(n-6)
A5n=n(n-1)…… (n-4)
So the problem is (N-5) (n-6) - 1 = 89
The solution is n = 15

Eight singers are going to participate in the festival and are going to arrange m performances for them. Four of them will perform on the stage each time. Any two of the eight singers are required to perform the same number of times at the same time. Please design a scheme to minimize the number of performances

This is a comprehensive problem of arrangement and combination. It's a little difficult, but it can be solved. First set the number of performances to m, then combine them, and finally arrange them. Alas, it's troublesome to input them. You can think about it carefully

A permutation and combination problem in Senior Two
What is the total number of plans for assigning anonymous volunteers to three different Olympic venues to participate in the reception work? What is the total number of plans for assigning at least one volunteer to each venue

In order to meet the requirements, five people need to be divided into three groups, namely 2 + 2 + 1, or 1 + 1 + 3, and then allocated to three venues
∴[(C5 2)(C3 2)/2+(C5 3)(C2 1)/2]×(A3 3)
=[10×3/2+10×2/2]×6
=(15+10)×6
=150

There are 9 red points and 5 yellow points on the plane. Among them, 2 red points and 2 yellow points are on the same straight line, and the others are not collinear. Take these points as vertices to make triangles. How many triangles are there with different colors of 3 vertices?
Brothers and sisters, I'm in urgent need. Please give me an explanation

The total number is C (14,3) - C (4,3) = 360
The coloring is exactly the same: C (9,3) = 84
Different colors: 360-84 = 276

On probability numbers
The ratio of the number of red balls to white balls in a bag is 1:9. If there are enough balls in the bag, now take out 3 balls randomly from the bag each time. What's the probability of taking out red balls twice in a row
How many red balls did you take out three times in a row,
How many times is four,
How many times is five
Sorry, enough in the title means that there are still balls to take after taking many times, and the ratio is constant 1:9 before taking the next time. It doesn't mean that if there are enough, that ratio can be ignored. What we should remember is that we can take three balls at a time, not only one,

"Enough" means that the number ratio of red ball to white ball is 1:9 before or after each win, so it is a Bernoulli model
The probability of each red ball is 1 - (9 / 10) ^ 3 = 281 / 1000, set as P
(1) P (continuous twice) = P × P = P & sup2;
(2) P (continuous 3 times) = P × P × P = P & sup3;
(3) P (continuous 4 times) = P &;
And so on