Extend the segment Ba to C reversely to make BC = 2 / 3AB, extend Ba to D to make Da = 1 / 3AB, DC = 6cm, and find the distance between the midpoint E and a of the segment DC Step by step, thank you

Extend the segment Ba to C reversely to make BC = 2 / 3AB, extend Ba to D to make Da = 1 / 3AB, DC = 6cm, and find the distance between the midpoint E and a of the segment DC Step by step, thank you

According to the meaning BC = 2 / 3AB, Da = 1 / 3AB
AD＝1/6CD,AB＝1/2CD,BC＝1/3CD
Since CD = 6
So ad = 1, de = 1 / 2CD = 3
EA＝DE－AD＝2

Extend the line segment AB to point C so that BC = 1 / 3AB, D is the midpoint of AC, and DC = 3cm, then what is the length of AB cm?

Let BC = X
Then AB = 3x
∴AC=4x
∵ DAC point CD = 6cm
∴AC=12cm
∴4x=12
x=3
∴AB=3x=9cm

Given the line segment AB, take a point C on the extension line of AB so that BC = 3AB, take a point D on the extension line of Ba so that Da = 32ab, e is the midpoint of DB, and EB = 30cm, and calculate the length of DC

As shown in the figure: ∵ Da = 32ab, e is the midpoint of DB, ∵ be = 54ab, be = 30cm, ∵ AB = 24cm, ∵ DC = BD + BC = 52ab + 3AB = 132cm

In the equal ratio sequence {an}, Sn is the sum of the first n terms. Let an > 0, A2 = 4, s4-a1 = 28. Find the value of a (n + 3) / an
Such as the title

S4=a1+a2+a3+a4
So s4-a1 = A2 + a3 + A4 = A2 + a2q + a2q ^ 2 = 28
a2=4
So 4 (1 + Q + Q ^ 2) = 28
q^2+q-6=0
(q+3)(q-2)=0
An is a positive number
So Q > 0
q=2
a(n+3)/an=q^3=8

If the general term of the equal ratio sequence {an} is an = 20-3n, note that the sum of the first n terms of {a} is Sn, and find SN

Let {an} be an arithmetic sequence, the first term A1 = 17, and the tolerance d = - 3. Just as the first floor said, there are two cases of summation, the second floor said only n ≤ 6, which does not hold when n ＞ 6: (1) when n ≤ 6, from the arithmetic sequence summation formula: SN = (17 + 20-3n) n / 2 = (37-3n) n / 2, so S6 = (37-3 * 6) 6 / 2 = 57 (...)

It is known that {an} is an equal ratio sequence of positive numbers, and A1 = 1, A2 + a3 = 6. Find the sum SN of the first 10 terms of the sequence

Let the common ratio of the sequence {an} be Q. from A1 = 1, A2 + a3 = 6, we can get: Q + Q2 = 6, that is, Q2 + q-6 = 0, and the solution is q = - 3 (rounding off) or q = 2  S10 = A1 (1 − Q10) 1 − q = 1 − 2101 − 2 = 210-1 = 1023

For the sequence {an}, if A1, a2a1, a3a2 ，anan−1，… Is an equal ratio sequence with the first term of 1 and the common ratio of 2, then A100 equals ()
A. 2100B. 299C. 25050D. 24950

According to the meaning: A100 = A1 × a2a1 × a3a2 × And A1, a2a1, a3a2 ，anan−1，… It is an equal ratio sequence with the first term of 1 and the common ratio of 2, that is, A1 = 1, a2a1 = 2, a3a2 = 22, Anan − 1 = 2n − 1, a100a99 = 299, A100 = A1 × a2a1 × a3a2 × ×a100a99=1×2×22×… ×299=2（1+2+… +99) and 1 + 2 + +99 = 4950  A100 = 24950, so the answer is: D

The common ratio of the equal ratio sequence {an} is Q, and the sum of the first n terms is SN
(1) If S5, S15, S10 are equal difference series, prove: 2s5, S10, s20-s10 are equal proportion series
(2) If 2s5, S10, s20-s10 are equal ratio series, please explain why S5, S15 and S10 are equal difference series

(1) When A1 is not equal to 0, 2s15 = S5 + S10 = > Q ^ 5 + Q ^ 10 = 2q ^ 15q = 1, and Q ≠ 1, 1 + Q ^ 5 = 2q ^ 10  2s5 (s20-s10) = 2A1 ^ 2 (Q ^ 10-2q ^ 15 + Q ^ 25) / (1-Q) & sup2; S10 & sup2; = A1 & sup2; (1-Q ^ 10) & sup2; / (1-Q) & sup2; (s20-s10) = S10 & sup2

(2012 · Anhui simulation) let the sum of the first n terms of the equal ratio sequence {an} be SN. & nbsp; if A1 = 1, S6 = 4S3, then A4 = ()
A. 2B. 3C. 4D. 5

In ∵ equal ratio sequence {an}, A1 = 1, S6 = 4S3, ∵ 1 × (1 − Q6) 1 − q = 4 × 1 × (1 − Q3) 1 − Q, the solution is Q3 = 3, ∵ A4 = 1 × Q3 = 3

In the equal ratio sequence {an}, Sn is the sum of the first n terms. Let an > 0, A2 = 4, s4-a1 = 28, find the value of A7 divided by A4!

Let the common ratio of equal ratio sequence {an} be Q
Because s4-a1 = 28
So A2 + a3 + A4 = 28
4 + 4q + 4q = 28
Q = 2 or q = - 3
Because let an > 0
So q = - 3
A7 divided by A4 = q to the negative third power = one eighth