As shown in the figure, point C is a point on line AB, m and N are the midpoint of line AC and CB respectively. Given AB = 10, find the length of Mn Question 2: as shown in the figure, point C is a point on line AB, and m and N are the midpoint of line AC and CB respectively. Given AB = α, find the length of Mn

As shown in the figure, point C is a point on line AB, m and N are the midpoint of line AC and CB respectively. Given AB = 10, find the length of Mn Question 2: as shown in the figure, point C is a point on line AB, and m and N are the midpoint of line AC and CB respectively. Given AB = α, find the length of Mn

MC=AM=AC/2
CN=BN=CB/2
CM+CN=AC/2+CB/2=(AC+CB)/2=AB/2=a/2
MN=CM+CN=a/2

As shown in the figure, P is the point on the fixed length line AB, and C and D respectively start from P and B and move to the left along the line AB at the speed of 1cm / s and 2cm / S (C is on the line AP and D is on the line BP)
(1) If there is always PD = 2Ac when C and d move to any time, please indicate the position of P on line ab
(2) Under the condition of (1), where q is a point on the straight line AB and aq-bq = PQ, the value of PQ / AB is obtained
(1) 2(AP -t ) =PB – 2t
2AP = BP P point is one third of a on line ab
(2) AQ=AP+PQ
BQ=BP-PQ
( AP+PQ )-( BP-PQ)=PQ
PQ=BP-AP=AB/3
PQ/AB=1/3

P is the point on the fixed line ab. C and d start from P and B and move to the left along the straight line AB at the speed of 1cm and 2cm per second respectively (C is on the line AP,
(1) if C and d move to any time, there will always be PD = 2Ac, find the quantitative relationship between line AP and line BP; (2) under the condition of (1), if point C continues to move to the left to the opposite extension line of line AB, and M is the midpoint of AC and N is the midpoint of BC, find the quantitative relationship between line Mn and line BP

Let AB = L, PA = x, Pb = LX, PC = 5, BD = 10, CD = PC + pb-bd = 5 + lx-10 = lx-5 = 1 / 2L, br / > × 5 = 1 / 2L, PA = x = 1 / 2l-5, Pb = 1 / 2L + 5, D continue to move, T seconds, CD = 1 / 2l-2t, PM = CD / 2-PC = 1 / 4l-t-7.5, PD = CD-5 = 1 / 2L

It is known that: as shown in Figure 1, M is a certain point on the fixed length line AB, C and D respectively start from m and B and move to the left along the straight line BA at the speed of 1cm / s and 3cm / s, and the direction of motion is shown by the arrow (C is on the line am, D is on the line BM)
(1) If AB = 10cm, when points c and d move for 2S, find the value of AC + MD. (2) if points c and d move, there is always MD = 3aC, fill in the blank directly: am=______ Ab. (3) under the condition of (2), where n is a point on the line AB and an-bn = Mn, the value of mnab is obtained

(1) When points c and d move for 2S, CM = 2cm, BD = 6cm ∵ AB = 10cm, CM = 2cm, BD = 6cm ∵ AC + MD = ab-cm-bd = 10-2-6 = 2cm (2) 14 (3) when point n is on line AB, as shown in the figure ∵ an-bn = Mn ∵ BN = am = 14ab, ∵ Mn = 12ab, that is, mnab = 12

It is known that the sum of the first n terms of an is SN. If the limit of limsn (n to positive infinity) is 1 / 4, what is the range of the first term A1

A1/(1-S) = 1/4
A1 = (1-S)/4
|S|

In the equal ratio sequence {an}, a 1 a 3 = 36, a 2 + a 4 = 60, Sn ＞ 400, find the range of n

According to the properties of equal ratio sequence, we can get that a1a3 = A22 = 36, A2 (1 + Q2) = 60, A2 > 0, A2 = 6, 1 + Q2 = 10, q = ± 3, when q = 3, A1 = 2, Sn = 2 (1 − 3n) 1 − 3 > 400, 3N > 401, n ≥ 6, n ∈ n; when q = - 3, A1 = − 2, Sn = − 2 [1 − (− 3) n] 1 − (− 3) > 400, (− 3) n > 8

In the equal ratio sequence (an), a1a3 = 36, A2 + A4 = 60, find A1, Q

If (an) is greater than 0
a1*a3=36
a3=a1*q^2
So A1 * A3 = A1 * A1 * q ^ 2 = A1 ^ 2 * q ^ 2 = 36
So A2 = A1 * q = 6
a2+a4=60
a2+a4=a1*q+a1*q^3=a1*q+a1*q*q^2
=6+6q=60
q=3
So A1 = 2

In the equal ratio sequence {an}, a 1 a 3 = 36, a 2 + a 4 = 60, Sn ＞ 400, find the range of n

According to the properties of equal ratio sequence, we can get that a1a3 = A22 = 36, A2 (1 + Q2) = 60, A2 > 0, A2 = 6, 1 + Q2 = 10, q = ± 3, when q = 3, A1 = 2, Sn = 2 (1 − 3n) 1 − 3 > 400, 3N > 401, n ≥ 6, n ∈ n; when q = - 3, A1 = − 2, Sn = − 2 [1 − (− 3) n] 1 − (− 3) > 400, (− 3) n > 8

In the equal ratio sequence {an}, a 1 a 3 = 36, a 2 + a 4 = 60, Sn ＞ 400, find the range of n

According to the properties of equal ratio sequence, we can get that a1a3 = A22 = 36, A2 (1 + Q2) = 60, A2 > 0, A2 = 6, 1 + Q2 = 10, q = ± 3, when q = 3, A1 = 2, Sn = 2 (1 − 3n) 1 − 3 > 400, 3N > 401, n ≥ 6, n ∈ n; when q = - 3, A1 = − 2, Sn = − 2 [1 − (− 3) n] 1 − (− 3) > 400, (− 3) n > 8

Given that A1 is equal to 1 in the equal ratio sequence, find the value range of S3

Let the common ratio be Q
a1=1
a2=a1*q=q
a3=a1*q²=q²
So S3 = 1 + Q + Q & # 178; = (Q + 1 / 2) &# 178; + 3 / 4
So when q = - 1 / 2
S3 min = 3 / 4
So the value range of S3 is [3 / 4, + ∞)