In the equal ratio sequence, "A2 > A4" is "A6 > A8"________ Condition () A. Sufficient and unnecessary B. necessary and not sufficient C. sufficient and necessary D. neither sufficient nor necessary

In the equal ratio sequence, "A2 > A4" is "A6 > A8"________ Condition () A. Sufficient and unnecessary B. necessary and not sufficient C. sufficient and necessary D. neither sufficient nor necessary

∵ A2 ∵ A4 ∵ sequence common ratio Q ≠ 0, ∵ Q4 ∵ 0 ∵ A6 = a2q4, a8 = a4q4, ∵ if A2 ∵ A4, there must be A6 ∵ A8, if A6 ∵ A8, there must be A2 ∵ A4, ∵ A2 ∵ A4 "is a necessary and sufficient condition for" A6 ∵ A8 ", so select C

If {an} satisfies A4 + A8 = - 3, then A6 (A2 + 2A6 + A10) = ()
A. 9B. 6C. 3D. -3

From the meaning of the question, we can get: in the equal ratio sequence {an}, if m, N, P, Q ∈ n *, and M + n = P + Q, then aman = apaq. Because A6 (A2 + 2A6 + A10) = a6a2 + 2a6a6 + a10a6, so a6a2 + 2a6a6 + a10a6 = (A4 + A8) 2 = 9

In the equal ratio sequence {an}, A5 = 162, common ratio q = 3, the first n terms and Sn = 242, find the first term A1 and the number of terms n

It is known that a1.34 = 162a1 (1 − 3n) 1 − 3 = 242, the solution is A1 = 2. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; substituting A1 = 2, we can get 2 (1 − 3n) 1 − 3 = 242, i.e. & nbsp; 3N = 243, the solution is & nbsp; & nbsp; & nbsp; n = 5. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp

In the equal ratio sequence {an}, A5 = 162, common ratio q = 3, the first n terms and Sn = 242, find the first term A1 and the number of terms n

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In the equal ratio sequence {an}, A5 = 162, common ratio q = 3, the first n terms and Sn = 242, find the first term A1 and the number of terms n

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In the equal ratio sequence {an}, A5 = 162, common ratio q = 3, the first n terms and Sn = 242, find the first term A1 and the number of terms n

It is known that a1.34 = 162a1 (1 − 3n) 1 − 3 = 242, the solution is A1 = 2. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; substituting A1 = 2, we can get 2 (1 − 3n) 1 − 3 = 242, i.e. & nbsp; 3N = 243, the solution is & nbsp; & nbsp; & nbsp; n = 5. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp

(1) If A1 = 9 / 8, q = 2 / 3, Sn = 65 / 24, then the number of terms n =?
(2) If a, B and C are equal difference sequence, and a, B, C + 3 and a + 1, B and C are equal ratio sequence, what are the numbers a, B and C?
(3) If LG10, y, lg10000 --- equal ratio sequence, then y =?

(1) Sn = A1 (1-Q ^ n) / (1-Q)} 9 / 8 × (1-Q ^ n) / (1-2 / 3) = 65 / 24} Q ^ n = (2 / 3) ^ n = 16 / 81 = (2 / 3) ^ 4} n = 4 (2)  a, B, C are equal difference sequence, and a, B, C + 3 and a + 1, B, C are equal proportion sequence  2B = a + C ① B &  178; = a (c + 3) ②

In the equal ratio sequence {an}, a1 + an = 34, A2 * a (n-1) = 64, the first n terms and Sn = 62. To find the value of the number of terms n and the common ratio Q needs a calculation process, but we can't work out thank you

a2*a(n-1)=a1*an=64
And a1 + an = 34
We can get A1 = 2, an = 32 or A1 = 32, an = 2
In the first case, Sn = A1 (1-Q ^ n) / (1-Q) = (a1-q * an) / (1-Q) = (2-32q) / (1-Q) = 62
We get q = 2
In this case, n = 5
The second one is q = 1 / 2, n = 5

(2011 · Sanya simulation) in the equal ratio sequence {an}, a1 + an = 34, A2 · an-1 = 64, and the first n terms and Sn = 62, then the number of terms n is equal to ()
A. 4B. 5C. 6D. 7

Because the sequence {an} is an equal ratio sequence, then A2 · an-1 = A1 · an = 64 ①, a1 + an = 34 ②, simultaneous ① ②, the solution is: A1 = 2, an = 32 or A1 = 32, an = 2, when A1 = 2, an = 32, Sn = A1 (1 − QN) 1 − q = (A1 − anq) 1 − q = 2 − 32q1 − q = 62, the solution is q = 2, so an = 2 × 2N-1 = 32, when n = 5; similarly, we can get A1 = 32, an = 2, also have n = 5. Then the number of terms n equals 5, so we choose B

In the equal ratio sequence {an}, a1 + an = 34, A2 times a (n-1) = 64, and the sum of the first n terms and Sn = 62, find the number of terms n

Then A2 * a (n-1) = A1 * an = 64
a1+an=34
Then A1 = 2, an = 32 or A1 = 32, an = 2
a1=2
an=a1*q^(n-1)=32
q^n=q*q^(n-1)=16q
Sn=2*(1-16q)/(1-q)=62
q=2
2^(n-1)=16=2^4
n=5
In the same way
a1=32,an=2
There is also n = 5
So n = 5