As shown in Figure 7, given the common part BC = 1 / 3AB = 1 / 4CD of line AB and CD, the midpoint of line AB and CD are m and N respectively, and Mn = 10cm, find the common part of line AB and CD

As shown in Figure 7, given the common part BC = 1 / 3AB = 1 / 4CD of line AB and CD, the midpoint of line AB and CD are m and N respectively, and Mn = 10cm, find the common part of line AB and CD

AB=12cm
CD=16cm
M is the midpoint of AB, so MC = 0.5ab-bc = 1.5bc-bc = 0.5bc
N is the midpoint of CD, so BN = 0.5cd-bc = 2bc-bc = BC
MN=MC+BN+BC=2.5BC=10cm,
So BC = 4cm,
AB=12cm,CD=16cm

As shown in the figure, ab = 20cm, points c and D are on the line AB, and CD = 8cm, m and N are the midpoint of AC and BD respectively, and the length of line Mn is calculated
.——.——.———.——.——
A M C D N B

It's not difficult. It's direct
AC+DB=AB-CD=12
So MC + DN = 12 / 2 = 6
So: Mn = MC + CD + DN = 8 + 6 = 14
OK, well

It is known that ab = 20 cm, CD = 8 cm, M is the midpoint of AD, n is the midpoint of BC, and the length of Mn is calculated. ② as shown in the figure, points E and F are the midpoint of line segments AC and BC respectively
EF = 2.5cm, find the length of line ab.

First question:
Let BC = a
Mn = md-nd = 1.2ad-nc-cd = 1 / 2 (AB + BC + CD) - nc-cd = 1 / 2 (20 + 8 + a) - 1 / 2a-8 = 14 + 1 / 2a-1 / 2a-8 = 6cm
Second question:
EF=EC-FC=1/2(AB+BC)-1/2BC=1/2AB+1/2BC-1/2BC=1/2AB
EF=2.5
AB=5

The sum of odd terms is 15. The sum of even terms is - 3
Find the first item

Let the first term be a and the ratio be Q
a/(1-q^2)=15
aq/(1-q^2)=-3
q=-1/5
a=125/8

The sequence an is an infinitely proportional sequence. The sum of all odd terms is 36, and the sum of even terms is 12. The general term formula of an is obtained

A1 of 1-Q is equal to 36, A2 of 1-Q is equal to 12

The first term is equal to 1, the sum of odd terms is 85, and the sum of even terms is 170
If the first term is equal to 1, the sum of odd terms is 85, and the sum of even terms is 170, find the common ratio and the number of terms of the sequence

If the number of items is even, the odd items are A1, A3, A5 The even number items are A2, A4, A6 Since the sum of all terms is 85 + 170 = 255 = A1 [1-Q ^ n] / (1-Q) = 255, the solution is n = 8 with 8 terms

If we know that the first term of an equal ratio sequence is 1, the number of terms is even, the sum of odd terms is 85, and the sum of even terms is 170, then the number of terms of this sequence is ()
A. 2B. 4C. 8D. 16

Let the common ratio be Q. from the meaning of the question, we can get a1 + a3 + +an-1=85，a2+a4+… +an=170，a1q+a2q+… +an-1q=170，∴（a1+a3+… +An-1) q = 170, q = 2, an = 2N-1, Sn = A1 (1 − QN) 1 − q = A1 − anq1 − Q, (Q ≠ 1) 170 + 85 = 2N-1, n = 8

The first term of a finite equal ratio sequence is 1, the number of terms is even, the sum of odd terms is 85, and the sum of even terms is 170. Find the common ratio Q?

n=2k
S odd = A1 (1-Q ^ 2K) / (1-Q)
S-even = A2 (1-Q ^ 2K) / (1-Q)
Q = A2 / A1 = s odd / s even = 2

If the first term of an equal ratio sequence with even terms is 1, the sum of odd terms is 85, and the sum of even terms is 170, then the common ratio of this sequence is? Detailed process, thank you

Analysis: let the equal ratio sequence be A1, A2, A3,..., a (n-1), an, where n is an even number and the common ratio is Q (Q ≠ 0), then it is defined by the equal ratio sequence (a sequence starts from the second term, and the ratio of each term to the previous term is equal to the same constant, then the sequence is equal ratio sequence). We can know: A2 / A1 = A4 / A3 = A6 / A5 =... = a (...)

An equal ratio sequence with 2n terms, prime minister is 1, and the sum of odd terms is 85. The sum of even terms is 170
seventy-nine

1 * (1-Q ^ 2n) / (1-Q ^ 2) = 85 and (Q * (1-Q ^ 2n)) / (1-Q ^ 2) = 170
The solution is q = 2
In sequence equation; (1-2 ^ (2n-1)) / (1-2) = 85 + 170 = 255
The solution is 2n = 9
So the common ratio is 2 and the number of terms is 9