In the arithmetic sequence {an} with non-zero tolerance, A3 = 7 and A2, A4, A9 are equal proportion sequence. (1) find the general term formula of sequence {an}. (2) let BN = 2An, find the first n term and Sn of sequence {BN}

In the arithmetic sequence {an} with non-zero tolerance, A3 = 7 and A2, A4, A9 are equal proportion sequence. (1) find the general term formula of sequence {an}. (2) let BN = 2An, find the first n term and Sn of sequence {BN}

(1) Let the tolerance of sequence be D, then ∵ A3 = 7, and A2, A4, A9 are equal ratio sequence. ∵ 7 + D) 2 = (7-d) (7 + 6D) ∵ D2 = 3D ∵ D ≠ 0 ∵ d = 3 ∵ an = 7 + (n-3) × 3 = 3n-2, that is, an = 3n-2; (2) ∵ BN = 2An, ∵ BN = 23n − 2 ∵ BN + 1bn = 23n + 123n − 2 = 8 ∵ sequence {BN} is equal
In the arithmetic sequence {an}, A4 = 10 and A3, A6, A10 are equal ratio sequence, find the sum of the first 20 terms of the sequence {an} S20
Let the tolerance of sequence {an} be D, then A3 = a4-d = 10-d, A6 = A4 + 2D = 10 + 2D, A10 = A4 + 6D = 10 + 6D. From A3, A6 and A10, a3a10 = A62, that is, (10-d) (10 + 6D) = (10 + 2D) 2, 10d2-10d = 0, and d = 0 or D = 1. When d = 0, S20 = 20a4 = 200
In the arithmetic sequence {an}, A4 = 10 and A3, A6, A10 are equal ratio sequence, find the sum of the first 20 terms of the sequence {an} S20
Let the tolerance of sequence {an} be D, then A3 = a4-d = 10-d, A6 = A4 + 2D = 10 + 2D, A10 = A4 + 6D = 10 + 6D. From A3, A6 and A10, we can get a3a10 = A62, that is, (10-d) (10 + 6D) = (10 + 2D) 2, then we can get 10d2-10d = 0, and get d = 0 or D = 1. When d = 0, S20 = 20a4 = 200. When d = 1, A1 = a4-3d = 10-3 × 1 = 7, then S20 = 20a1 + 20 × 192D = 20 × 7 + 190 = 330
Log (a ^ n) M =? For example, log (2 ^ 2) 3 =?
log(a^n)M=1/n*（logaM）,
Log(2^2)3=1/2log（2）3
In the circuit shown in the figure, the power supply voltage is 3V. When the switches S1 and S2 are closed, then ()
A. The indication of voltmeter is 3vb. Lamp L1 is not on, lamp L2 is on. C. ammeter will be burned out. D. both bulbs will be burned out
It can be seen from the circuit diagram that when the switches S1 and S2 are closed, the two bulbs and the voltmeter are short circuited, the current starts from the positive pole, passes through the ammeter, and directly returns to the negative pole of the power supply, ∵ when the electric appliance is short circuited, it can not work, ∵ the two bulbs are not on and will not be burned out, and the voltmeter has no indication, so the abd is incorrect, ∵ when the power supply is short circuited, the current is too large, it will burn out the power supply and the ammeter, ∵ the current is too The watch will be burnt out, so C is correct
Prove that log (a) (m · n) = log (a) m + log (a) n
It is proved that if log (a) (MN) = k, then a ^ k = Mn
Log (a) M = x is a ^ x = M
Log (a) n = y is a ^ y = n
We can get: Mn = a ^ x * a ^ y = a ^ (x + y) = a ^ K
That is: k = x + y
The result is: log (a) (MN) = log (a) m + log (a) n
Let the a power of a be m and the B power of a be n, then Mn = a + B power of a, log (a) (MN) = a + B = log (a) m + log (a) n
log(a) (M·N)=log(a) M+log(a) N
a^x=M, ,,,,,,(1) a^y=N, ..........(2)
Change (1) (2) into the logarithmic formula x = log (a), y = Mlog (a) n
(1) * (2) a ^ x · a ^ y = m · n
It can be expressed by logarithm
a^(x+y)=M·N
log(a) (M·N)=x+y=log(a) M+log(a) N
Let m = a ^ P, n = a ^ Q
log(a) (M·N)=log(a) a^(p+q)=p+q
log(a) M+log(a) N=p+q
∴ log(a) (M·N)=log(a) M+log(a) N
In the circuit as shown in the figure, the voltage at both ends of the power supply remains unchanged. When the switch S1 is closed, the change of ammeter and voltage indication after the switch S2 is closed is compared with that when the switch S2 is open
A. A voltmeter with a small number of ammeters is a voltmeter with a large number of ammeters
(1) The equivalent circuit diagram when both switches S1 and S2 are closed is shown in figure a, and the equivalent circuit diagram when switch S1 is closed and switch S2 is open is shown in Figure B. (2) according to the equivalent circuit diagram, after switch S2 is closed (as shown in figure a), compared with when switch S2 is open (as shown in Figure b), the voltmeter always measures the voltage at both ends of the power supply
Log (a) M = m, log (a) n = n, how to get log (a) (m × n) = m + n
According to the logarithm algorithm: log (a) (m * n) = log (a) m + log (a) n
log (a)(M*N)=log a (M)+loga(N)=m+n
Isn't this the formula "log (a) m + log (a) n = log (a) (m * n)"
As shown in the figure, the center angle of the circle is 30 ° and the radii are 1, 3, 5, 7 The area of shadow part is S1, S2, S3 , then S14=______ π was retained
S1=30π×32360-30π×12360=8π12，S2=30π×72360-30π×52360=8π12×3，S3=30π×112360-30π×92360=8π12×5，S4=8π12×7，… According to the meaning of the question, we can get the general formula: SN = 8 π 12 × (2n-1), that is, Sn = 8 π 12 × (2n-1),  S14 = 23 π (28-1) = 18 π, so the answer is: 18 π
It is known that ab = m (a > 0, b > 0, M is not equal to 1), and log (m) (b) = x, Ze log (m) (a) is equal to_____
log(M) ab = log (M) M =1
Moreover, log (m) AB = log (m) a + log (m) B = log (m) a + X
So log (m) (a) = 1 - X