How many questions should I ask about the proportional series, and how to do this proportional serie

1.Knowing that x, 2x+2, 3x+3 is a proportional sequence, the first three terms are what is the fourth term?

According to the nature of the proportional series:

2x+2）^2=x*(3x+3)

4x^2+8x+4=3x^2+3x

x^2+5x+4=0

x=-1 (rounded, because according to high school regulations, the image of a proportional series cannot be 0) or x=-4

So x=-4 the fourth term is.

2.If three unequal real numbers a, b, and c are in a series of equal differences, and a, c, and b are in a series of proportional numbers, find a:b:c

The real numbers a, b, and c are in a series of equal differences: a+c=2b.

c=2b-a

a, c, b are proportional to the series ab=c 2

So (a-2b) 2=ab

a^2-5ab+4b^2=0

a-b)(a-4b)=0

Because a, b, and c are unequal real numbers.

So a-4b=0

a=4b c=2b-a=-2b

a：b：c =4：1：（-2）

3.Knowing that the common ratio of the proportional series is 2, and the product of the first ten terms is 50 of 2, find a2=a1*2

a4=a2*2^2

a6=a3*2^3

a20=a10*2^10

So please seniors Be sure to write the steps clearly Little brother, I have a poor foundation, I can't write the steps clearly, I can't understand it, thank you.

Solution: 1, x, 2x+2, 3x+3 is a proportional series, so there is (2x+2) 2=x(3x+3), both sides of the equation are divided by x+1 at the same time, and 4(x+1)=3x, which gives x=-4, so these three terms are: -4, -6, -9, and the fourth term is:

2. a, b, c into a series of equal differences, that is, 2b=a+c; a, c, b are proportional sequences, i.e., c 2 = ab.

So c 2 = a(a + c) 2, 2c 2 = a 2 + ac, and both sides of the equation are divided by a 2 (the square of a) at the same time, and we get:

1+c a-2c 2 a 2=0, treat c a as an unknown number, and solve c a=1 or -1 2, because the problem has said that it is an unequal real number, so it is equal to 1 does not fit the meaning of the question, then c a=-1 2, that is: a=-2c, b=-c 2, so a:b:

c =-2c：(-c/2)：c=-2：

3. From the known: q=2, s10=2 to the 50th power, s10=a1(1-q 10) (1-q)=a1(2 10-1)=2 50

So a1=2 50 1023, then a2=a1*2=2 51 1023, a3=a2*2......And so on.

1, (2x+2) x=(3x+3) (2x+2) gives x=-1 or -4, where x=-1 is not desirable, and the first three terms of this ratio series are -4, -6, -9, then the fourth term is.

2. Because the real numbers a, b, and c are equal difference series, and a, c, and b are equal to the proportional series, there are ab=c 2, 2b=a+c, and they are jointly solved by substituting c=2b-a and a=2b-c respectively to obtain a:b=4, b:c=, so a:

b：c=8：2：

13. The question is incomplete or you missed the word yourself???

If it is a sum, then a1 + 2a1 + 4a1 + 8a1 + 16a1 + 32a1 + 64a1 + 128a1 + 256a1 + 512a1 = 2 50 50 to get a1 = 2 50 1023, and the answer will be calculated slowly.

If it is a product, there is a1*2 (1+2+3+4+5+6+7+8+9)=2 50, and we get a1=2 5, a2=2 6

So the answer = a2*2 (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18) = 2 96

1.(2x+2) 2 = x*(3x+3), the solution gives x = -1 or x = -4, when x = -1, 2x+2 = 0, so rounded, x = -4, the first three terms are -4, -6, -9, so the fourth term is.

2. 2b = a+c;c 2 = ab The solution is a = 2c, b = 1 2c, so a:b:c = 2:(1 2):1 = 4:1:2

3.You typed one word less, is it the sum or the product of the first fifty items?

Yes and: no solution.

is the product: an = 2,4,6....

a2*a4*..a20 = 2*2^1*2*2^3*..2*2^19 = 4*4*4^1*..4*4^18 = 4^19*4^((0+18)*19/2) = 4^132

1, x*(3x+3)=(2x+2)*(2x+2) gives x=-1, or x=-4

When x=-1, the second term of the original sequence is 2x+2=-2+2=0, which is no longer proportional to the series, so it is rounded.

x=-4, the original series -4, -6, -9, the fourth term -27 22, the difference series, the difference is k, b = a + k, c = a + 2k proportional series, a*b = c*c, so there is a*(a+k)=(a+2k) 2, and the solution is a=-4k 3

So b=-k 3, c = 2k 3

a:b:c=(-4 3):(1 3):(2 3)=-4:-1:23, I understand that the "and" of the first ten terms is 50 of 2.

s10=a1(1-2 10) (-1)=2 50 solves a1 and then you can find it.

1. From (2x+2) 2=x*(3x+3), x=-4 (the other value -1 does not exist, so it is rounded), so that the fourth phase is.

2. A=C is obtained from b 2=ac and 2b = a+c, and thus b = a=c = any real number (except 0).

3. Write the formula of the sequence an=2 (n-1)*a1, the first 10 are equal to 50 to calculate a1=2 43, so that a2=4 43, a4=2 4 43, a6=2 6 43, a8=2 8 43, a10=2 10 43, and so on, until a20 is equal to 2 to the 20th power ratio of 43

1. (2x+2)2=x(30+3), the solution is x=-1, or -4, and substitution can be known to round -1, so the first three terms are -4, -6, -9, and the fourth term is -7 2.

2. From the question, it can be seen that 2b=a+c, c2=ab, 2c2=a2+ac, that is, a2+ac-2c2=0, (a+2c)(a-c)=0 solution: a=-2c or a=c (rounded), substituted to obtain c=-2b, so a:

b:c=-4:-1:

2。3. Is it the sum of the first ten terms to the 50th power of 2? Sum or product? Gotta be clear.

1. Find x 4 and bring in 1 2 (9x 9).

2. A+C=2B and AC B 2, and there is ABC!=0, abc is not equal to each other! The solution yields a=4b, c=-2b

So: a:b:c 4:1:2

3. There is a formula, forget it, eat and do it again.

an=a3,a2,a4 in a series of equal differences.

a3+a4=2a2

q^2+q^3 =2q

q(q^2+q-2)=0

q(q+2)(q-1)=0

q=-2an = 2^ns5

bn= -1+(n-1)d

b2, b3-1, b4 in proportional series.

b3-1)^2

1+d)(-1+3d) =2+2d)^23d^2-4d+1 = 4d^2 -8d+4d^2-4d+3=0

d-1)(d-3)=0

d=1 or 3

bn= -1+(n-1) or -1+3(n-1)=n-2 or 3n-4

t10= 35 or 125

1. The pre-meaning of the envy elimination of the proportional series:

In general, if the ratio of each term to its predecessor from the second term of a series of numbers is equal to a constant (not 0), then the sequence is called the proportional number brother clear column, and this constant is called the common ratio of the equal bridge ratio sequence, which is usually expressed by q.

The definition can be expressed by the formula: a(n+1) an=q (where n is a positive integer and q is a constant). It is important to note that q is a constant independent of the number n.

2. Equal ratio of the middle term:

The three numbers a, g, and b form a proportional sequence in turn, then g is called the proportional middle term, and g2=a+b (the square of the proportional middle term is equal to the product of the previous term and the next term).

2. The relevant formulas of the proportional series:

1. The general term formula of the proportional series: When talking about the general term formula of the equal difference series, the basic method of solving the general term formula of the number series was introduced - the accumulation method, and today we introduce another basic method for solving the general term formula of the number series - the cumulative multiplication method.

Multiply both sides of the equation by a1 at the same time to get the formula for the general term of the proportional series.

Accumulation and accumulation are the two basic methods for solving the general term formula of the series, and mastering it is of great help to learn the recursive method to solve the general term formula of the series, so we should think about it carefully.

a1+a1q=-1①

a1-a1q =-3 , obtained by: (1 ten q) (1-q )=1 3, q +3q + 2 = 0, q = -1 or q = -2, and the remainder is stupid.

When q=-1, there is no solution, when q=-2, and attack a1=1, then a4=a1q calls the destroyer brother=-8, in summary: a4=-8.

-3=-1 3 (to the right of the equation is a 3x macro segment).

a1-a1q = (a1+a1q) 3=3a1+3a1q (the left side of the equation is 3 times).

OK. Or all in the form of AQ.

A4=AQ 3,A6=AQ 5, then A 2Q 6=AQ 5, AQ=1,A2=1

Correct, because in the proportional series, if.

m ten n = p ten q, then am·an=ap·aq.

a4) = a4·a4 = a2·a6, a2·a6 = a6, in proportional series, a6 ≠0, a2 = 1.

Proof: because (3-m)s

n+2man=m+3

1 so (3-m) s (n-1).

2ma(n-1)=m+3

2 Formula 1 - Formula 2 has.

3-m)an+2m(an-a(n-1))=0, i.e. (3+m)an=2mA(n-1).

So) = 2m (3+m).

Because m is not -3 and m is not 0, an a(n-1) is a constant that is not 0 and an is a proportional series.

ps: applied to sn-s(n-1)=an, (n-1) after a and s in the process of solving the problem is the subscript,

Proportional series Let the ratio be q

a1=1a8=1*q^7=128 q=2

So these six numbers are 2 4. 8.

16, and is =126a3=2s2+1(@)a4=2s3+1( ) and the equation is subtracted left and right -@=a4-a3=2(s3-s2)=2a3, so a4=3a3 q=3

One, it is obvious that 128 = 2 7, and the common ratio is 2, so 6 numbers are 2, 4, 8 ......64, which adds up to 126

Second, subtract the first one by the second formula, and we get a4-a3=2a3, so a4=3a3, so q=3.

Good guaranteed

1.Let the ratio be d. Then 1*d 7=128

D=2 is obtained, so the 6 numbers are respectively.

And for: 126

2.According to a3=2s2+1, a4=2s3+1, s2=(a3-1) 2, s3=(a4-1) 2, and s3=s2+a3

then we get a4=3a3

Therefore, the common ratio q is equal to 3

a(n+1)=2sn+1

s(n+1)=sn+a(n+1)=3sn+1 formula, s(n+1)+(1 2)=3[sn+(1 2)] makes cn=sn+(1 2), cn is a proportional series of 3.

c1=s1+(1/2)=a1+(1/2)=3/2cn=(1/2)*3^n

sn=(1 晌sailsilver2)*3 n-(1 2)an=sn-s(n-1)=3 (n-1), which is an equal proportional series.

There is the above banquet, a1 = 1, a2 = 3, a3 = 9 is the equal difference series and the car is a positive number, b1>0, tolerance d 0b2 = b1 + d, b3 = b1 + 2d

a1 + b1, a2 + b2, a3 + b3 in proportional series.

a1+b1)(a3+b3)=(a2+b2) substitution, simplified, 4b1=d +4d ..Type 1.

t3=153b1+3d=15 ..Type 2.

It is solved by equation 1 and equation 2, b1 = 3, d = 2

bn=2n+1,tn=n²+2n