It is known that there is a point C on the line AB, M is the midpoint of AC, n is the midpoint of BC, ab = 12 cm. Please draw a figure according to the title and find out the length of Mn Please write down the process clearly, thank you

It is known that there is a point C on the line AB, M is the midpoint of AC, n is the midpoint of BC, ab = 12 cm. Please draw a figure according to the title and find out the length of Mn Please write down the process clearly, thank you

Cm = 1 / 2Ac, CN = 1 / 2BC, CM + CN = Mn = 1 / 2Ac + 1 / 2BC = 1 / 2 (AC + BC) = 1 / 2Ab = 6cm

As shown in the figure, C is any point of the line AB, M is the midpoint of AC, and N is the midpoint of ab. if BC is equal to 10 cm, find Mn

∴AN=(AC+BC)/2,AM=1/2AC
Ψ Mn = an-am = (AC + BC) / 2-ac / 2 = BC / 2 = 5cm

As shown in the figure, CD is two points on the line AB, CD is equal to eight centimeters, M is the midpoint of AC, Mn is equal to twelve centimeters, so how many centimeters is the line AB equal to?

Kiss us go on

In the equal ratio sequence {an}, it is known that A3 = 3 / 2, S3 = 9 / 2, and the values of A1 and Q are obtained

According to the meaning of the title: a1 + A2 + a3 = 9 / 2, x0da3 = 3 / 2, x0da1 + A2 = 6 / 2, x0da1 = 3-a2, x0d (3-a2) * 3 / 2 = A2 ^ 2, x0da2 = 3 / 2. If A2 = 3, then A1 = 3 / 2, q = 1

If A3 = 3 / 2 and S3 = 9 / 2 are known in the sequence of equal ratio numbers, find A1 and Q

A1 * q ^ 2 = A3 = 3 / 2A1 (1-Q ^ 3) / (1-Q) = S3 = 9 / 2 grain division Q ^ 2 (1-Q) / (1-Q ^ 3) = 1 / 33q ^ 2 (1-Q) = (1-Q ^ 3) 2q ^ 3-3q ^ 2 + 1 = 02q ^ 2 (Q-1) - (Q-1) (Q + 1) = 0, and Q ≠ 12q ^ 2-q-1 = 0 (2q + 1) (Q-1) = 0, and Q ≠ 12q + 1 = 0, q = - 1 / 2A1 = A3 / Q ^ 2 = 6, so A1 = 6, q = - 1 / 2

(3) The sum of the first n terms of the equal ratio sequence {an} is SN. Given that S3 = A2 + 10A1, A5 = 9, find A1
We should take concrete steps

S3=a1(1+q+q²)=a1(q+10)
1+q+q²=q+10
q²=9
q=±3
a5=a1×q^4=a1×81=9
Then A1 = 1 / 9

If S3 = A2 + 10A1, A5 = 9, then A1 =?

Solution
s3=a2+10a1
Namely
a1+a2+a3=a2+10a1
∴a3=9a1
∴q²=a3/a1=9
∴q²=9
∵a5=a1q^4=9
∴a1=9/9²=1/9

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

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

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