Given the line AB = 10cm, C is on the line AB, M is the midpoint of AC, n is the midpoint of BC, 1. When C is on the line AB, find Mn 2. When C is on the extension of Ba Finding the length of Mn

Given the line AB = 10cm, C is on the line AB, M is the midpoint of AC, n is the midpoint of BC, 1. When C is on the line AB, find Mn 2. When C is on the extension of Ba Finding the length of Mn

∵ C is on the extension line of the line AB, ab = 10, BC = 4 ∵ AC = 14 ∵ m is the midpoint of AC ∵ am = 7 cm, there is another way to get three, the result of this problem is 3 or 7, because AB = 10, ab-cb = 6,

It is known that all items of the equal ratio sequence an are positive. A1 = 2. The sum of the first three items is 14. The general term formula of an is obtained. Let BN = log2an, the sum of the first 20 items of BN is obtained

Let the common ratio of an equal ratio sequence be Q
From A1 = 2, the sum of the first three terms is 14, 2 + 2q + 2q ^ 2 = 14
The solution is q = - 3 (rounding off) or 2, that is, the general formula with an is an = 2 ^ n
And BN = log2an, so BN = n sequence {BN} is an arithmetic sequence with 1 as the first term and 1 as the tolerance
Let the sum of the first n terms of the sequence {BN} be Sn, then: S20 = [20x (1 + 20)] / 2
=210

It is known that all the items of {an} are positive numbers, A1 = 2, and the sum of the first three items is 14. The general term formula of {an} is obtained

In {an}, A3 = 7, the first three terms and S3 = 21, then q is the common ratio=______ ．

From A3 = 7, S3 = 21, we can get: a1q2 = 7a1 (1 + Q) = 14, we can get q = - 0.5 or 1, so the answer is - 0.5 or 1

In {an}, A3 = 7, the first three terms and S3 = 21, then q is the common ratio=______ ．

From A3 = 7, S3 = 21, we can get: a1q2 = 7a1 (1 + Q) = 14, we can get q = - 0.5 or 1, so the answer is - 0.5 or 1

It is known that the first term of the equal ratio sequence {an} is A1 and the common ratio is Q. if n is even, then the N / 2 term is

a(n/2)=a1*q^(n/2-1)

In known equal ratio sequence {an}, A1 = 2, S3 = 26. Find Q and A3

S3=a1+a2+a3=26
a2+a3=24
a1q+a1q²=24
So Q & # 178; + q = 24 / A1 = 12
q²+q-12=0
q=-4,q=3
q=-4,a3=a1q²=32
q=3,a3=a1q²=18

If A3 = 7, the first three terms and S3 = 21, then what is the value of common ratio q,
a3=a1*q^2=7; (1)
s3=a1(1-q^3）/（1-q)=a1(1+q+q^2)=21; (2)
(2) (1) get
Q = 1, or q = - 1 / 2;
Why A1 (1-Q ^ 3) / (1-Q) = A1 (1 + Q + Q ^ 2) = 21?

This is factorization
1-q^3=(1-q)(1+q+q^2)

It is known that the first term A1 of the equal ratio sequence {an} is 1, and the common ratio 0 〈 Q 〈 1. Let the general term of the sequence {BN} be BN = an + 1 + an + 2, then the general term formula of the sequence {BN}
N + 1 is a term, not an plus 1

An=q(n-1),An+1=qn,An+2=q(n+1),bn=qn(1+q)

It is known that the sequence {an} is an equal ratio sequence whose common ratio is greater than 1. For any n ∈ n *, an + 1 = a1 + A2 +... + an-1 + 5 / 2An + 1 / 2
1. Find the general term formula of sequence {an}
2. Let the sequence {BN} satisfy: BN = 1 / N (log3 (A1) + log3 (A2) +... + log3 (an) + log3 (T)) (n ∈ n *), if {BN} is an arithmetic sequence, find the value of real number T and the general term formula of the sequence {BN}

an=a1.q^(n-1)a(n+1)=a1+a2+...+a(n-1)+(5/2)an+1/2 n=1,a2=(5/2)a1+1/2a1q =(5/2)a1+1/2 (1)n=2,a3=a1+(5/2)a2+1/2a1q^2=a1+(5/2)a1q+1/2 (2)(2)-(1)q(q-1) =-(3/2) +(5/2)q2q^2-7q+3=0(2q-1)(q-3)=0q=3 (q>1)from ...