If DC = 39cm, find the length of line AB (write process: because, so)

If DC = 39cm, find the length of line AB (write process: because, so)

Let AB = X
Then BC = 1 / 3x, ad = 1 / 9x
DC=DA+AB+BC=1/9x+x+1/3x=13/9x=39cm
AB=X=27cm

Given the line segment AB, extend AB to point C to make BC = AB; extend Ba to point d to make Da = 3AB, then DC = () AB, AC = () BC, AC = () BC
DA＝（）DB?

If line AB is known, extend AB to point C to make BC = AB; if Ba to point d to make Da = 3AB, then DC = (5) AB, AC = (2) BC, Da = (3 / 4) dB

If the line segment AB is known, extend the line segment AB to point C so that BC = 1 / 3AB, extend the line segment AC to point D in reverse so that Da = 1 / 2Ac, if BC = 3, then DC = several centimeters

BC=1/3AB BC=3
So AB = 9
So AC = 9 + 3 = 12
DA=1/2AC=12x1/2=6
DC=DA+AC=6+12=18

It is known that each item of the equal ratio sequence {an} is a positive number, Sn = 80, s2n = 6560, and in the first n terms, the largest term is 54. Find the value of n

If q = 1, then there is a contradiction between Sn = Na1 = 80, s2n = 2Na1 = 160 and s2n = 6560, so Q ≠ 1. ∵ A1 (1 − QN) 1 − q = 80 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (1) & nbsp; A1 (1 − q2n) 1 − q = 6560 & nbsp; & nbsp; (2) In the first n terms, the largest term is an, that is, an = 54. And an = a1qn-1 = a1qqn = 54, and QN = 81, ｜ A1 = 5481q. That is, A1 = 23q. Substituting A1 = 23q into (1), we get 23q (1-qn) = 80 (1-Q), that is, 23q (1-81) = 80 (1-Q), and the solution is q = 3

In the equal ratio sequence {an}, A1 > 0, Sn = 80, s2n = 6560, the term with the largest median value of the first n terms is 54, and the general term formula an is obtained
(a1-a1*q^n)/(1-q)=80
(a1-a1*q^2n)/(1-q)=6560
I don't know how to get these two steps

According to the meaning of the title: A1 > 0, Q > 1,
a1*q^(n-1)=54
(a1-a1*q^n)/(1-q)=80
(a1-a1*q^2n)/(1-q)=6560
Divide the two: 80 / 6560 = 1 / [1 + Q ^ n] = > Q ^ n = 81
Bring Q ^ n into
The solution is: A1 = 2; q = 3; n = 4

In the equal ratio sequence (an), A1 = 3, an = 96, Sn = 189, find the value of common ratio Q and item number n

An=A1×q^(n-1)=3q^(n-1)=96
q^(n-1)=32
S(n-1)=Sn-An=189-96=93
S(n-1)=A1×(1-q^(n-1))/(1-q)
=3(1-32)/(1-q)=93
q=2
2^(n-1)=32
n=6

In the equal ratio sequence {an}, A1 = 3, an = 96, Sn = 189, how much is n

According to the formula:
a1*q^(n-1)=an=96;
a1(1-q^n)/1-q=189;
The solution is q = 2;
n=6.

In the known equal ratio sequence {an}, an = 96, Sn = 189. Q = 2, find the sum of N and a

Sn = [A1 anq] / [1-Q], 2an-a1 = 189, A1 = 3, so an = 3 × 2 ^ (n-1), from an = 96, n = 6

If the common ratio q = 2, an = 96 and the sum of all terms Sn = 189 of the equal ratio sequence {an} are known, then the sequence has the same number__ First item A1=
cry for help..

If Sn = 189, an = 96, q = 2, then A1 =?
rt

Use the formula
sn=(a1-an*q)/(1-q)
Substituting data
So A1 = 3