It is known that A1 = 64 in the arithmetic sequence an, common ratio q is not equal to 1, A2, A3, A4 are the 7th, 3rd and 1st terms of the arithmetic sequence respectively 1. Find an 2. Let BN = log2 find the first n terms of {BN} and TN Why use common ratio Q to express A2, A3 and A4, and then use the mean term of arithmetic to get the solution q = 1?

Let the common ratio of {an} be Q. let A2 = 64q, A3 = 64q & # 178;, A4 = 64q & # 179; and because A2, A3 and A4 are the 7th, 3rd and 1st terms of the arithmetic sequence respectively, then let {CN} be the arithmetic sequence, the first term of which is C1, and the tolerance is dc7 = C1 + 6D
I love mathematics + love numbers, love me = to learn (in vertical)
For example, in addition calculation, the same Chinese character represents the same number, and different Chinese characters represent different numbers. It is known that "Xi" = 5. To make this formula true, what are the five digits represented by "I love learning mathematics"?
It is known that the number of pieces in the box is between 130 and 160. If one piece is taken out of the box for the first time and two pieces are taken out for the second time, and so on, there are still nine pieces left, which is not enough to be taken again, then how many pieces are there in the box/
The number of chessmen in the box is between 130 and 160. Take one chessman out of the box for the first time, and take two for the second time. And so on, there are still nine chessmen left. How many chessmen are there in the box? 160-1-2-3-4. - 17 = 7. So, no more than 17 chessmen, 1 + to 17 equals 153163 minus 153 equals 9
97 out of 99 times 100 simple operation! My sixth grade, as simple as possible! Everyone help me!
97 / 99 * 100
=97 out of 99 * (99 + 1)
=97 * 99 of 99 + 97 * 1 of 99
=97 / 99
9700 / 99 = 97.9798 by calculator
97/99x100
=97/99x(99+1)
=97/99x99+97/99
=97+97/99
=97 and 97 / 99
97 out of 99 = 1 minus 2 out of 99,
Then use (1 minus 2 / 99) X100,
That's 100 minus 200 out of 99,
200 out of 99 = 2 and 2 out of 99,
So 100 minus 99 / 200 = 97 and 97 / 99.
I think we can write 100 as 99 + 1, 97 / 99 x (99 + 1) = 97 and 97 / 99
It is known that A1 = 64 in the arithmetic sequence an, common ratio q is not equal to 1A2, A3 and A4 are the 7th, 3rd and 1st terms of the arithmetic sequence respectively
Find an
On the absolute value of the sum of logbn = 2
(a1*q-a1*q^2)/(7-3)=(a1*q^2-a1*q^3)/(3-1)
So q = 1 / 2
Let the arithmetic sequence be CN
c1=8
cn=4n+4
an=2^(7-n)
bn=log2(an)=7-n
When n = 7, TN = - (6 + 7-N) n / 2 + 2 * (6 + 0) 7 / 2 = - (13-n) n / 2 + 42
Love me to learn multiplied by love me to learn is love mathematics love mathematics
What do these Chinese characters stand for?
Love me to learn * love me to learn = love math, love math
Love = 1, I = 7, learning = 5, number = 2
Love = 2, I = 8, learning = 6, number = 3
175*715=125125
286*826=236236
Simple calculation method of 47 / 99 × 97 91 / 100 × 99
47/99×99-47/99×3
91/100×100-91/100×1
In the known arithmetic sequence {an}, A2, A3 and A4 are the 5th, 3rd and 2nd terms of an arithmetic sequence respectively, and A1 = 12, common ratio Q ≠ 1. (1) find the general term formula of the sequence {an}; (2) the known sequence {BN} satisfies: A1B1 + a2b2 + +Anbn = 2N-1 (n ∈ n *), find the first n terms and Sn of sequence {BN}
(1) According to the known condition, A2-A3 = 2 (a3-a4), that is, A1 (q-q2) = 2A1 (q2-q3), the solution is 2q3-3q2 + q = 0, and the solution is q = 12 or q = 1 (rounding off) or q = 0 (rounding off), so an = (12) n. (2) when n = 1, A1B1 = 1, ｜ B1 = 2, when n ≥ 2, A1B1 + a2b2 + + an-1bn-1 + anbn = 2N-1 (...)
In the following multiplication formula, what does each man represent? Math paradise * 9 = happy learning numbers
Mathematics paradise
×………… Nine
Happy learning mathematics
Four digits multiplied by 9 or four digits
The number is 1 and the garden is 9
1 Xuele 9
×……… Nine
9 happy learning 1
Similarly, learning cannot be greater than 1, it can only be 1 or 0
Substituting verification, 1 does not conform
Learning is 0, music is 8
The number of mathematics paradise is 1089, and the number of happy learning is 9801
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A and B trucks of a logistics company start from a and B which are 300 km apart and run in opposite directions at the same speed. When the two trucks run for 1.5 hours, car a first arrives at the distribution station C. at this time, the distance between the two trucks is 30 km. Car a takes | hours to distribute goods in place C, and then drives to place B according to the original speed. Car B also drives to place C for 2 hours, and continues to drive to place a without stopping, B. What's the distance between C and a? What's the distance between a and C? What's the speed of car a and the time it takes for car a to reach B? How long does car B leave? The distance between the two cars is 150 km
Speed of car B: 30 / 2 = 15 (km / h) BC distance: 15 * (1.5) + 30 = 52.5 (km) AC distance: 300-52.5 = 247.5 (km) speed of car a: 247.5 / 1.5 = 165 (km / h) time to B: 1.5 + 1 + 52.5 / 165 = 31 / 11 (H) distance before the first meeting: 150 km
The answers are as follows:
Speed of vehicle a and B: 30 / 0.5 = 60km / h
Distance between B and BC: 60x2 = 120km
Distance between C and AC: 300-120 = 180km
Speed of car D A: 180 / 1.5 = 120km / h
Arrival time of e a vehicle at B: 1.5 + 1 + (120 / 180) * 60 = 3 h 10 min
How long does it take for car F B to leave? The distance between the two cars is 150km (150-60x0.5) / (60 + 120) X60 = 40min + 30min = 1h10min
In the known sequence {an}, A1, A2 and A3 are equal difference sequence, A2, A3 and A4 are equal ratio sequence, and the reciprocal of A3, A4 and A5 are equal difference sequence,
In the known sequence {an}, A1, A2, A3 are equal difference sequence, A2, A3, A4 are equal ratio sequence, and the reciprocal of A3, A4, A5 are equal difference sequence. What sequence are A1, A3, A5
In proportion
The arithmetic of equal difference and equal ratio
Just replace it
First of all, it seems that your question is incomplete, but it does not hinder the answer
If you think about it simply, it must be special. You can guess that the five a's are equal
The steps are as follows
Equation 1 2A2 = a1 + a3
Equation 2 a2a4 = (A3)^
Equation 3 2 / A4 = 1 / A3 + 1 / A5
If you multiply formula 2 by formula 3, you get 2A2 = A3 + (A3) ^ / A5. If you want to subtract Formula 1, you get (A3) ^ = a1a5
Expansion of equation 3
First of all, it seems that your question is incomplete, but it does not hinder the answer
If you think about it simply, it must be special. You can guess that the five a's are equal
The steps are as follows
Equation 1 2A2 = a1 + a3
Equation 2 a2a4 = (A3)^
Equation 3 2 / A4 = 1 / A3 + 1 / A5
If you multiply formula 2 by formula 3, you get 2A2 = A3 + (A3) ^ / A5. If you want to subtract Formula 1, you get (A3) ^ = a1a5
Equation 3 divides a1a5 + A3A5 = A4 (A3 + A5), then use (A3) ^ = a1a5 to get A3 (A3 + A5) = A4 (A3 + A5)
It is obtained that A3 = A4 is brought into formula 2, A2 = A3 is brought into Formula 1, A1 = A2 is brought into formula 5, A4 = A5 can be obtained
Every a of the synthesized sequence is equal

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