In the equal ratio sequence {an}, a1 + A6 = 33, A3 · A4 = 32, a (n + 1)

In the equal ratio sequence {an}, a1 + A6 = 33, A3 · A4 = 32, a (n + 1)

a1+a1q^5=a1(1+q^5)=33
a1q^2*a1q^3=a1^2*q^5=32
So A1 = 32, q = 1 / 2
Or A1 = 1, q = 2
Because a (n + 1)

In the equal ratio sequence an, a1 + A6 = 33, A3 * A4 = 32, an
N-1 power of 1.2
2.n（n-1）/2

a1+a6=33
a1(1+q^5)=33
a3*a4=32
(a1)^2*q^5=32
q^5=32/(a1)^2
a1[1+32/(a1)^2]=33
a1+32/a1=33
(a1)^2+32=33a1
(a1)^2-33a1+32=0
(a1-32)(a1-1)=0
A1 = 32 or A1 = 1
an

Given an equal ratio sequence {an}, a1 + a3 = 10, A4 + A6 = 54, find the first five terms and S5

Let the common ratio of the equal ratio sequence {an} be Q. from the meaning of the question, we can get a1 + a3 = A1 (1 + Q2) = 10, ① A4 + A6 = a1q3 (1 + Q2) = 54, ② division of the two formulas, ② ① can get Q3 = 18, the solution can get q = 12, and substituting into ① can get A1 = 8, ｛ S5 = A1 (1 − Q5) 1 − q = 8 (1 − 125) 1 − 12 = 312

In the equal ratio sequence {an}, a1 + a3 = 10, A4 + A6 = 4 / 5, find S5

(a4+a6)/(a1+a3)=q^3=5/4/10=1/8
q=1/2
a1+a1/4=10
a1=8
S5=15.5

Given sequence {an} satisfies A1, a2-a1, A2-A3, a4-a3 , an-a (n-1) is the expression of finding an from the equal ratio sequence (1) with Prime Minister of 1 and common ratio of 1 / 3

A2-a1 = 1 / 3, a3-a2 = (1 / 3) ^ 2... An-a (n-1) = (1 / 3) ^ (n-1) n ≥ 2, an-a1 = 1 / 2 * [1 - (1 / 3) ^ (n-1)] an = 1 / 2 * [1 - (1 / 3) ^ (n-1)] + 1 = 1 / 2 * [3 - (1 / 3) ^ (n-1)] n ≥ 2, n = 1, an = 1 satisfies the above formula, so an = 1 / 2 * [3 - (1 / 3) ^ (n-1)]

In the plane rectangular coordinate system xoy, a point m on the hyperbola x24 − y212 = 1, and the abscissa of point m is 3, then the distance from m to the right focus of the hyperbola is______ ．

If MFD = e = 2, D is the distance from point m to the right guide line x = 1, then d = 2, MF = 4

In the plane rectangular coordinate system xoy, if the eccentricity of hyperbola x2m − y2m2 + 4 = 1 is 5, then the value of M is 0______ ．

Therefore, A2 = m > 0, B2 = M2 + 4, C2 = m + M2 + 4 = M2 + m + 4, the eccentricity of hyperbola x2m − y2m2 + 4 = 1 is 5, CA = 5, C2 = 5A2, so M2 + m + 4 = 5m, the solution is m = 2, so the answer is: 2

In the plane rectangular coordinate system xoy, if the eccentricity of hyperbola x2m − y2m2 + 4 = 1 is 5, then the value of M is 0______ ．

Therefore, A2 = m > 0, B2 = M2 + 4, C2 = m + M2 + 4 = M2 + m + 4, the eccentricity of hyperbola x2m − y2m2 + 4 = 1 is 5, CA = 5, C2 = 5A2, so M2 + m + 4 = 5m, the solution is m = 2, so the answer is: 2

It is known that the hyperbola x2 / 9-y2 / 27 = 1 and m (5,3) f are the right focus. If there is a point P on the hyperbola to minimize PM + 1 / 2 PF, then the coordinate of point P is?
Why X / X1 = 1 / 3

1
Hyperbola x2 / 9-y2 / 27 = 1
a²=9,b²=27,c²=a²+b²=36
c=6,a=3,e=c/a=2
Right focus f (6,0), right collimator L: x = A & # / C = 3 / 2
PN ⊥ l over P is better than N, MQ ⊥ l over M is better than Q
According to the second definition of hyperbola:
|PF|/|PN|=e=2
∴1/2|PF|=|PN|
∴|PM|+1/2 |PF|
=|PM|+|PN|≥|MQ|=5-3/2=7/2
If and only if m, P and N are collinear, the equal sign is taken
The ordinate of point P is 3,
x²/9-9/27=1
Ψ X & # 178; = 12, x = 2 √ 3 (rounding off)
∴P（2√3,3）
two
I'm not clear about C1C2. The topic is not clear

If the distance from a point P on the hyperbola x24 − Y22 = 1 to the right focus of the hyperbola is 2, then the distance from point P to the Y axis is ()
A. 463B. 263C. 26D. 23

The distance from the point P to the right focus (6,0) of the hyperbola is 2, and we know that P is on the right branch of the hyperbola. From the second definition of the hyperbola, we know that the distance from the point P to the right quasilinear of the hyperbola is 263, and the right quasilinear equation of the hyperbola is x = 263, so the distance from the point P to the Y axis is 463