High school sophomore series problem, very urgent, help me

Because lg2, lg(2 x -1), lg(2 x+3) are equal difference series.

Then: LG2+LG(2 x+3)=2LG(2 x -1)LG2(2 x+3)=LG(2 x -1) 2 gets: 2(2 x+3)=(2 x -1) 2Let a=2 x, then the above equation can be reduced to:

2(a+3)=a^2-2

a^2-2a-8=0

a-4)(a+2)=0

Get a=4 or a=-2

Swap into the original formula, because 2 x cannot be less than 0, then 2 x=4 can be obtained, x=2

Because the sum of the first n terms of the proportional series is 2 n -1, then.

a1=s1=1 a2=s2-s1=2 a3=s3-s2=4 common ratio q=2

an=2^(n-1)

Then there is a series of numbers where an 2=[2 (n-1)] 2=4 (n-1) shows that the sequence is a proportional sequence with 1 as the initial number and the common ratio q is 4, which can be obtained from its summing formula, and the sum of the series sn=a1(1-q n) (1-q)=(4 n-1) 3

4: (Not so sure).

Let the three sides be a, b, and c in length, and since the three sides are proportional series, then there is.

When the triangle is an equilateral sequence, then q=1

When the triangle is not an equilateral sequence, let ac

Again, b=a*q c=a*q 2, substitution can be obtained, a+a*q>a*q 21+q>q 2

q^2-q-1<0

According to the basic formula of the analytic formula of the quadratic function, it can be obtained.

The axis of symmetry is q=1 2, and the distance between the two points is 1 2*root number 14, because q is the common ratio of the three sides, which must be greater than 0

Therefore, the value range of q (0, 1 2+1 4, root number 14), this range already contains 1.

a1+a1*q*q*q=18 (1)

a1*q+a1*q*q=12 (2)

Substituting (2) from (1) gets: a1=18 (1+q*q*q) into (2): 3*q+3*q*q=2+2*q*q*q3*q*(1+q)=2*(1+q)*(1-q+q*q)2*q*q-5*q+2=0

Get q = 2 or 1 2

When q=2, a1=2

When Q=1 2, A1=16

Questions 1, 2, 3, and 5 are correct.

In question 4, q can take +2 and -2, and you missed -2.

Problem 6 is miscalculated, s3=2(1+q+q2)=26, solve q=3 or -4, and the rest of an can be brought into the solution by yourself.

Question 7 because an is a proportional sequence, so a1a3=(a2) 2, bring in a1a2a3=8 to get a2=2, and we know a1+a2+a3=-3, so 2 q+2+2q=-3, the solution is q=-1 2 or -2, so a4 will naturally be found, and I will not count it.

Question 8: a5a9=(a7) 2, bring in the data to get a9=9

Question 9 a5-a1=a1(q 4-1)=15,a4-a2=a1(q 3-q)=6, divide the two formulas, then eliminate a1, get (q 2+1) q=5 2, so you can solve q, and then bring in any of the above formulas, you can find a1, natural a3=a1*(q 2), ask for it yourself, I don't ask for it.

Question 10 I can't understand your answer, my answer is q 5 = a9 a4 = 243, I get q = 3, so an=4*[3 (n-4)].

Question 11 is the same as question 10, first seek the common ratio Q, calculate it yourself, I won't count it anymore, if you really won't ask me again.

This topic is so confusing, how can I wait for the difference, and I can't figure out which one is the difference and which is the same.

Since Question 12 gives linear recursion, why is it still said to be a series of equal differences? If you only use the eigenequation to solve the condition of the recursive relation, you shouldn't be in high school yet, right? The eigenequation is a method in the math competition, which can solve all linear recursive equations, and you can study it if you are interested, but I won't talk about it here, it's too tiring.

Question 13 is obviously positive for the first 12 items, 0 for the 13th item, and the rest are all negative, and the biggest one is s13 with common sense.

Hungry, what kind of number sequence is question 14, I can't think of how to do it with such conditions.

If I don't write it, I'm almost exhausted, these questions are very basic number series problems, and I guess you haven't learned these things yet, so it may be difficult to do; I am a graduate of the third year of high school this year, anyway, the advice to you is to memorize more than 20 formulas related to the sum and equal difference, equal ratio and sum, and then master the basic methods, as well as the method of finding the general term of the characteristic root, then it is more than enough to cope with the college entrance examination.

If the above topic is still really unclear, my QQ is 873650215

1.The proportion of the area reclaimed this year to the existing tidal flat area is x

The total area of the final reclamation land shall not exceed the following

x + 95x+(.95)^2x+..= x * 1 (so 20x <= 20%, x<=1%.)

This year, the area reclaimed will only account for a maximum of 1 percent of the existing tidal flats

2.The area of reclamation is x

ax^2 + xb <= x* 10b

ax <= 9b

x <= 9b/a

So the maximum value of the reclamation area x is 9b a

(1) If the problem is not complete, at least give a few years to complete it, or what year it is this year.

2) Funding ax 2 + average annual economic income before reclamation bx< = 10bx

That is, ax 2-9bx>=0 x(ax-9b)>=0 is defined by the question x must be "0,a>0,b>0, then x>=9b a