Line BD = 1 / 3AB = 1 / 4CD, points m and N are the middle points of line AB and CD respectively, and Mn = 20cm, find the length of AC A -- m -- D -- B -- N -- C

48cm
If BD is regarded as one part, the total AC is 6 parts and Mn is 2.5 parts. Then divide 20 by 2.5 times 6 to get 48CM

Given that the sequence {an} is an equal ratio sequence with the first term of 1, Sn is the sum of the first n terms of {an}, and 9s3 = S6, then the sum of the first 5 terms of {an} is ()
A. 30B. 31C. 29D. 32

According to the meaning of the title, the common ratio Q ≠ 1, ｜ 9 · 1 − Q31 − q = 1 − q61 − Q, that is, q6-9q3 + 8 = 0, the solution is: q = 2 or q = 1 (rounding off), ｜ S5 = 1 − 251 − 2 = 31, so B

It is known that (an) is an equal ratio sequence with the first term of 1, Sn is the sum of the first n terms of (an), and 9s3 = S6, then the sum of the first 5 terms of an is ()

9S3=S6
9（1-q³）/（1-q）=(1-q^6)/(1-q)
q^6-9q³+8=0
（q³）²-9q²+8=0
（q³-1）（q³-8）=0
Q & # 179; = 1 (rounding off) or Q & # 179; = 8
So q = 2
S5=(1-2^5)/(1-2)=31
Ask again if you don't understand

In the equal ratio sequence {an} with even terms and positive terms, the sum of all its terms is equal to 4 times of the sum of even terms,
A 2 a 4 = 9 (a 3 + a 4). Find a 1 and the common ratio Q. find the value of n when the first n terms of the sequence {lgan} are maximized

The sum of all its terms is equal to four times the sum of even terms
Even term sum / odd term sum = q = 1 / 3
a2a4=9（a3+a4)
a3^2=9a3+9a3q
a3=12
a1=a3/q^2=108
an=a1*q^(n-1)=108/3^(n-1)=4/3^(n-4)
n1 lgan>0
n>5
an

The number of terms of the positive term sequence {an} is even. The sum of all its terms is equal to 4 times of the sum of its even terms. The product of the second and fourth terms is 9 times of the sum of the third and fourth terms
Problem (1) finding A1 and common ratio Q
(2) The first few terms of {lgan} and the largest?

Odd term * q = even term
So odd + even = even / Q + even = 4 * even
The solution is q = 1 / 3
a2*a4=a2*a2*q^2=1/9*(a2^2)
a3+a4=a2(1/3+1/9)=a2*4/9
So: A2 = 0 or A2 = 4
a1=12;
lgan=lg(12*(1/3)^n)=lg12-n*lg(3)
Make lgan > 0
The average score was more than 1
The solution is n

It is known that the number of terms of a positive proportional sequence {an} is even, the sum of all its terms is equal to 4 times of the sum of its even terms, and the product of the second term and the fourth term is zero
9 times of the sum of the third term and the fourth term to find the general term formula of the sequence {an}

If the number of terms of an is even, and the sum of all its terms is equal to 4 times of the sum of its even terms, then the sum of odd terms is 3 times of the sum of even terms. Therefore, the common ratio of an is 1 / 3a2 * A4 = 9 (A3 + A4) A1 * A1 * q ^ 4 = 9a1q & # 178; (1 + Q) A1 = 9 * (1 + Q) / Q & # 178; = 81 (1 + 1 / 3) = 108 an = 108 *

The sum of all the terms of an equal ratio sequence with even terms is equal to 4 times of the sum of its even terms, and the product of the second and fourth terms is 9 times of the sum of the third and fourth terms
Find the general term formula of the sequence

Note that the common ratio is Q and the sum of even terms is a
Sum of even terms = q × sum of odd terms
S = a + A / Q = 4A, q = 1 / 3
a2*a4=a1^2*q^4=9(a3+a4)=9(a1q^2+a1q^3)
The result is: A1 = 9 (1 + Q) / Q ^ 2 = 108
a[n]=108*(1/3)^(n-1)

In the equal ratio sequence {an} with positive items, it is known that the number of items is even, and the sum of all items is equal to 4 times of the sum of even items, and a2a4 = (A3 + A4) * 9
(1) Finding A1 and Q
(2) Find the n value when the sum of the first n terms of the sequence {lgan} is the maximum

(1) Let an = A1 * q ^ (n-1), then s = A1 * (1-Q ^ n) / (1-Q) = sum of even terms of A1 * [1-Q ^ (2k)] / (1-Q), s (even) = A1 * Q * [1-Q ^ (2k)] / (1-Q ^ 2) because s = 4 * s (even) A1 * [1-Q ^ (2k)] / (1-Q) = 4 * A1 * Q * [1-Q ^ (2k)] / (1-Q ^ 2), q = 1 / 3, because a2a4 = (A3 + A4) * 9a1 * q * a

The equal ratio sequence an, A1 = 1 has 2K terms, the sum of odd terms is 85, and the sum of even terms is 170. Then the common ratio q is
Why did others calculate q = 2?

Is the sign of odd and even items the same in the equal ratio sequence?
Why can different sequences be real numbers? If they are complex numbers???? For example, A2 = 2, A6 = 8. A4 =? Is it positive or negative 4? Q square = positive or negative 4, the square formula of complex I - 1, how to explain
I want to ask
Is the sign of odd and even items in the equal ratio sequence the same.. Not the sign of odd and even terms...
After learning the complex number, is it odd? The symbols of each item can be different

All the odd numbers have the same sign
All even numbers have the same sign

Line BD = 1 / 3AB = 1 / 4CD, points m and N are the middle points of line AB and CD respectively, and Mn = 20cm, find the length of AC A -- m -- D -- B -- N -- C