As shown in the figure, the line AB = 8 cm, m and N are the midpoint of BC and AC respectively, then what is the length of Mn, On the same plane

As shown in the figure, the line AB = 8 cm, m and N are the midpoint of BC and AC respectively, then what is the length of Mn, On the same plane

four
If C is not on AB, ABC forms a triangle, Mn is ab, is the median line, Mn = 1 / 2Ab = 4
If C is on AB, CM = 1 / 2MB, CN = 1 / 2mc, so Mn = 1 / 2Ab = 4

If M is the midpoint of AC, n is the midpoint of BC, and ab = 10cm, the length of Mn is obtained

AB = 10, so BC = 20 BN = 10 BM = 15 AB = 10 BM = 5 Mn = BN BM = 5

C is the point on the line AB, AC = 10cm, M is the midpoint of AB, n is the midpoint of CB, find the length of Mn

MN=MB+BN
=1/2AB+1/2BC
=1/2(AB+BC)
=1/2AC
=5cm

If A1 = 1 in the equal ratio sequence {an}, then the value range of sum S3 of the first three terms is______ ．

Let the common ratio of the equal ratio sequence {an} be q (Q ≠ 0), A1 = 1, A2 = a1q = q, A3 = a1q2 = Q2, and S3 = a1 + A2 + a3 = 1 + Q + Q2 = (q-12) 2 + 34. If q = 12, S3 has the minimum value, and the minimum value is 34, then the value range of sum S3 of the first three terms is [34, + ∞]. Therefore, the answer is: [34, + ∞)

If the sum of the first three terms of the arithmetic sequence {an} is S3 = 9 and A1 = 1, then A2 equals ()
A. 3B. 4C. 5D. 6

∵ S3 = 9 and A1 = 1, ∵ S3 = 3A1 + 3D = 3 + 3D = 9, the solution is d = 2. ∵ A2 = a1 + D = 3

If the sum of the first n terms of an is Sn, and S3 = (S2) &# 178;, then the value range of the first term A1 is

From the problem we know that an / a (n-1) = k
So A2 = KA1 A3 = ka2 = K & # 178; A1
S3=a3+a2+a1=k²a1+ka1+a1=S2²=(a2+a1)²=(ka1)²+2ka1²+a1²=a1²(k²+2k+1)
therefore
a1=(k²+k+1)/(k²+2k+1)=1-k/(k²+2k+1)=1-1/(k+2+1/k)
Because K + 1 / k > = 2
So K + 2 + 1 / k > = 4
1-1/(k+2+1/k)0
So 0

If A2 = a (a > 0) in the equal ratio sequence {an}, then the value range of sum S3 of the first three terms is

Let the common ratio be Q
a1 = a2/q = a/q
a3 = a2*q
S3 = a(1 + 1/q + q)
= (a/q) * (q^2 + q + 1)
= (a/q) * (q^2 + q +1/4 + 3/4)
= (a/q) * [(q + 1/2)^2 + 3/4]
So S3 and Q have the same number
If you have learned derivative, you can further:
S3‘ = a(1 - 1/q^2)
When 1 - 1 / Q ^ 2 = 0, there is an extremum
Then q = ± 1
S3’‘ = 2a/q^3
When Q > 0
S3'' > 0
When q = 1, there is a minimum
S3(1) = 3a
When Q < 0
S3" < 0
There is a maximum when q = - 1
S3(-1) = -a

In sequence an, A1 = 1, a (n + 1) = (an + 2) / an, and BN = (An-2) / (an + 1), (1) prove that BN is an equal ratio sequence; (2) find Sn and limsn of BN

1) B (n + 1) = [a (n + 1) - 2] / [a (n + 1) + 1] = [(an + 2) / An-2] / [(an + 2) / an + 1] = [an + 2-2an] / [an + 2 + an] = (2-An) / (2 + 2An) = - 1 / 2 * (An-2) / (an + 1) = - 1 / 2 * BN

If the common ratio of the equal ratio sequence {an} is - 1 / 2, the first n terms and Sn satisfy limsn = 1 / A1, then the first term A1

The sum of the first n items of the sequence is:
Sn＝a1(1-q^n)/(1-q)
Because q = - 1 / 2, limn - > + ∞ Sn = 1 / A1
So LIM (n - > + ∞) Sn
=lim(n->+∞) a1(1-q^n)/(1-q)
=a1/(1+1/2)
=1/a1
Then (A1) &# 178; = 3 / 2
The solution is A1 = √ 6 / 2 or A1 = - √ 6 / 2

It is known that the sum of the first n terms of a sequence an composed of positive numbers is SN
If A1 = 1 / 2, and N is the median of Sn and 1 / an, find limsn

Because: n is the middle term of the equal ratio of Sn and 1 / an, so: n & sup2; = Sn / an, Sn = n & sup2; an; s (n-1) = (n-1) & sup2; a (n-1); by subtracting the two formulas: SN-S (n-1) = n & sup2; an - (n-1) & sup2; a (n-1); so; an = n & sup2; an - (n-1) & sup2; a (n-1), the term is reduced to: (n + 1) an = (n-1) a (n-1)