Senior 1 Mathematics Proportional Number Series

Let the first term = A1 be the common ratio to q

s7=[a1(1-q^7)]/(1-q)

s14-s7=[a1(q^7-q^14)]/(1-q)

s21-s14=[a1(q^14-q^21)]/(1-q)

s14-s7)^2=[a1^2(q^14-2q^21+q^28)]/(1-q)^2

s7*(s21-s14)=[a1^2*(q^14-2q^21+q^18)]/(1-q)^2

So(s14-s7) 2=s7*(s21-s14).

So s7, s14-s7, s21-s14 are proportional sequences.

sk=[a1(1-q^k)]/(1-q)

s2k-sk=[a1(1-q^2k)-a1(1-q^k)]/(1-q)

s3k-s2k=[a1(1-q^3k)-a1(1-q^2k)]/(1-q)

sk(s3k-s2k)=(s2k-sk)^2

So sk, s2k-sk, s3k-s2k are proportional series.

The proportional sequence is d i.e. ak=d*a(k-1).

then s2k-sk=d k (because a(k+s)=d k * a(s)).

a3=a1*q 2=e (b2)=e 18a6=a1*q 5=e (b6)=e 12, then a6 a3=q 3=e 12 e 18=e (-6) gives q=e (-2), a1=e 22

The general formula for proportional series is e (24-2n).

The sequence satisfies bn=ln(an).

Then the general formula for the series is (24-2n).

The upstairs prompt is correct, but the b12 term is discarded, and the sum of the first 11 terms is still 132) When n=12, bn=0, and an is not equal to 1, then bn=0 does not exist, so there is no b12 term.

But when n>12, bn < 0

So when the first n terms and sn are taken as the maximum, n = 11

Then sn(n=11)=(b1+b11)*11 2=132

Because it is a proportional series, a3 a1=a5 a3= a4 a2=a6 a4, so (a5+a6) (a3+a4)=(a3+a4) (a1+a2)

So (a5+a6) = (a3+a4)*(a3+a4) (a1+a2).

You can substitute yourself for the rest of the one.

a3+a4=(a1+a2) multiplied by q squared; (a3+a4) divided by (a1+a2)=q squared = 4

a5 + a6 = (a1 + a2) times q to the fourth power = 30 times 4 squared = 480

2x,3y,4z in proportional series, 9y 2=8xz 1 equation 1 x,1 y,1 z 2 y=1 x+1 z squared 2 y 2=1 x 2+2 xz+1 z 2 2 equation.

1 by 2.

18=8z/x+16+8x/z

8z/x+8x/z=4

x/z+z/x=1/2

d if a>0

an=a*q^(n-1)

Because an is an increasing sequence.

So q (n-1) > 1

q>1 if a<0

Because an is an increasing sequence.

So 00

a(n+1)+1=2a(n)+1+1=2(an+1), the latter term is twice as large as the previous term, a(1)+1=1+1=2, the prime minister is not 0, so it is an equal proportional series.

a(n)+1} is a proportional sequence with a prime of 2 and a common ratio of 2, so.

a(n)+1=2^n

So. a(n)=2^n-1

No, because the situation of q=1 is unlikely (a3>s3), you can directly use the formula to represent a3 and s3 with a1 and q.

Let the first item A1 be compared to Q

A1-1, A1Q-1, A1Q 2-4, A1Q 3-14, into the same liquid hand difference resistance series.

a1q-1+a1q 2-4=a1-1+a1q 3-14 gives a1==3 q=2

The four numbers are 3, 6, 12, and 24

If a, b, and c are in an equal difference series, and a, c, and b are in an equal proportional series, then the value of a b is ( ).

The original question is wrong! It should not be a proportional sequence of a, c, and d, but a proportional sequence of a, c, and b, otherwise it cannot be solved. )

Solution: a, b, c into a series of equal differences, a+c=2b....1)

and a, c, b in proportional sequences, c = ab....2)

C=2b-a is obtained by (1) and substituted by equation (2) obtains: (2b-a) =ab

That is, there is b [2-(a b)] = ab, so [2-(a b)] = a b, that is, there is (a b) -4(a b)+4=a b, a b) -5(a b)+4=0

a/b)-1][(a/b)-4]=0

a b) = 1 (rounded), or (a b) = 4

Discussion: If a b=1, then there is a=b=c, where a,b,c are constant sequences with a tolerance of 0; a, c, and b are proportional sequences with a common ratio of 1, so they should be discarded.

If a b=4, then a=4b, c=-2b; 4b, b, -2b are equal difference sequences with a tolerance of -3b, while 4b, -2b, and b are a common ratio of -1 2

proportional series.

The title is wrong, a, c, b are proportional series.

Tolerance of the equal difference series = (c-b) 2

Common ratio of proportional series = under the root sign (b a).

2b=a+c

c = under the root number (ab).

2b-a = under the root number (ab)->a 2-4ab-4b 2=ab(a-b)(a-4b)=0

So a b = 1 or 4

I suspect you should have written the title incorrectly, is it a, c, b in proportional numbers, right? If so, the solution is as follows:

Two equations can be listed: 2b=a+c, c 2=a b, (2b-a) 2=a b, there is a=b or a=4b; then the result of a b is 1 or 4.

According to the title, 2b = a + c, the square of c = ab, get c = 2b-a, and the square of the two sides will be squared into c to get (b-a) (4b-a) = 0, so as to get a b = 1 or 4

Not a ......

If you have to do the math, a=c2 d

b=(c+c2/d)/2

a/b=(c2/d)/((c+c2/d)/2)=2c/(c+d)

a, b, c in a series of equal differences, 2b = a + c c = 2b-a, a, c, b in a proportional series, c = ab (2b-a,) = ab 4b -4ab + a = ab

4b -5ab+a =0 solution, b=a, 4b=a understand?

Let's change the title for you, otherwise how to do it.

If a, b, and c are in a series of equal differences, and a, c, and b are in a proportional series, then the value of a b is changed from ( ) d to b

a, b, c into a series of equal differences.

2b=a+c (1)a,c, b in proportional series.

b = ac (2) Eliminate c to get a -5ab + 4b =0

a/b)²-5(a/b)+4=0

a b=4 a b=1 (rounded).

If a, b, and c are in a series of equal differences, and a, c, and b are in an equal proportional series, then the value of a b is (4).