In the equal ratio sequence an, TN is the product of the first n terms. If TN = 1, then A .a2=1 B.a3=1 C.a5=1 D.a9=1

In the equal ratio sequence an, TN is the product of the first n terms. If TN = 1, then A .a2=1 B.a3=1 C.a5=1 D.a9=1

Because an is an equal sequence
So TN is also an isometric sequence
Tn=An*A(n-1)*A(n-2)*```*A1
Because an = A1 * q ^ (n-1)
So TN = A1 ^ n * q ^ (1 + 2 + 3 + +n)
Tn/T(n-1)=A1*q^n
A1=T1
Tn=T1*(A1*q^n)^(n-1)=A1*(A1*q^n)^(n-1)=A1^n*q^(n(n-1))
A1^n*q^(n(n-1))=A1^n*q^(1+2+3+… +n)
That is, n (n-1) = 1 + 2 + 3 + N
Replace 2359 into
When n = 3, left = right
So A3 = 1

If the first term A1 = 3 and the sum of the first three terms is 21, then A3 + A4 + A5 = () A. 33 B. 72 C. 84 D. 189

In the equal ratio sequence {an}, the first term A1 = 3 and the sum of the first three terms is 21
So 3 + 3Q + 3q2 = 21,
∴q=2，
∴a3+a4+a5=（a1+a2+a3）q2=21×22=84
Therefore, C

It is known that {an} is an equal ratio sequence of positive numbers, and a1 + A2 = 2 (1 / A1 + 1 / A2), A3 + A4 + A5 = 64 (1 / A3 + 1 / A4 + 1 / A5) The general term formula of {an}

Let the common ratio be Q
a1+a2=2(1/a1+1/a2)=>a1(1+q)=(2/a1q)*(q+1)=>a1^2*q=2
a3+a4+a5=64(1/a3+1/a4+1/a5)=>a3(q^2+q+1)=64/(a3*q^2)(q^2+q+1)=>(a3*q)^2=a1^2*q^6=64
Because {an} is positive, A4 = A3 * q = 8
And Q ^ 5 = 64 / 2 = 32, q = 2
So A1 = 1, an = 2 ^ (n-1)

If the first term A1 = 3 and the sum of the first three terms is 21, then A3 + A4 + A5 = () A. 33 B. 72 C. 84 D. 189

In the equal ratio sequence {an}, the first term A1 = 3 and the sum of the first three terms is 21
So 3 + 3Q + 3q2 = 21,
∴q=2，
∴a3+a4+a5=（a1+a2+a3）q2=21×22=84
Therefore, C

Let BN = 1 / (an) ^ 2-1 (n ∈ n *), Find the first n terms and TN of sequence {BN}

If the first item is A1 and the tolerance is D, then a1 + 2D = 7, a1 + 4D + A1 + 6D + 6D = 26, the solution is A1 = 3, d = 2, so an = a1 + (n-1) d = 2n + 1, then BN = 1 / [(an) ^ 2-1] = 1 / [4N (n + 1)] = 1 / 4 * [1 / n-1 / (n + 1)], so TN = B1 + B2 +. + BN = 1 / 4 * [1-1 / 2 + 1 / 2-1 / 3 + 1 / 3-1 / 3 + 3-1 / 3-1 / 3-1 / 4 +. [1-1 / 2 + 1 / 2-1 / 3 + 1 / 3-1 / 4 +. [1-1 / 3 / 3-1 / 3-1 / 4 +. + 4+ 1 / n-1 / (n + 1)]

It is known that the arithmetic sequence {an} satisfies: A3 = 7, A5 + A7 = 26. The sum of the first n terms of {an} is SN. Let BN = 1 / (an) ^ 2-1, find {BN} and the sum of the first n terms TN

a3=7
a5+a7=2a6=26
a6=13
a6-a3=6=5d-2d=3d,d=2
a1+2d=7=a1+4 a1=3
an=3+2(n-1)=2n+1
bn=1/[an^2-1]=1/[4n(n+1)]=(1/4)(1/n-1/(n+1))
b1=(1/4)(1-1/2)=1/8
Tn=(1/4)(1-1/(n+1))

It is known that the arithmetic sequence {an} satisfies: A3 = 7, A5 + A7 = 26. The sum of the first n terms of {an} is SN (1) Find an and Sn (2) Let BN = 1 / (an squared) - 1 (n is contained in N +), and find the first n terms and TN of the sequence BN

As shown in the figure:

Let {an} satisfy: A3 = 7, A5 + A7 = 26, the sum of the first n terms of {an} is SN. Find an and Sn

Let the tolerance of arithmetic sequence {an} be d,
be
a3=a1+2d=7
a5+a7=2a1+10d=26 ，
The solution
a1=3
d=2 ，
∴an=3+2（n-1）=2n+1
Sn=n(3+2n+1)
2=n2+2n

We know the first n terms and Sn of the arithmetic sequence {an} with tolerance greater than zero, and satisfy A3 * A4 = 117, A2 + A5 = 22 Find: 1. Arithmetic sequence {an} 2. If the sequence {BN} is an arithmetic sequence, BN = Sn / (n + C), find the nonzero constant C; 3. The maximum value of F (n) = BN / [(n + 36) BN + 1] (n ∈ n +)

Because an is an arithmetic sequence with tolerance d > 0,
So A2 + A5 = 22 = A3 + A4
a3*a4=117
So we get A3 = 9, A4 = 13
So the tolerance d = a4-a3 = 13-9 = 4
So A1 = 1
1）、an=a1+（n-1）*d=1+（n-1）*4=4n-3
2）、Sn=（a1+an）*n/2=（1+4n-3）*n/2=n（2n-1）
So BN = n (2n-1) / (n + C) is an arithmetic sequence, and C ≠ 0
Then n has no quadratic term, so C = - 0.5
3、bn=2n
f（n）=2n/〔（n+36）*2（n+1）〕=1/（n+37+36/n）≤1/（37+2√36）=1/7
When n = 36 / N and N = 6, f (n) max = f (6) = 1 / 7

It is known that the sum of the first n terms of the arithmetic sequence {an} is Sn, and a1 + a3 + A5 = 105, A2 + A4 + A6 = 99, then Sn obtains the maximum value of n=______ .

∵a1+a3+a5=105，a2+a4+a6=99，
∴3a3=105，3a4=99，∴a3=35，a4=33
The tolerance d = - 2
∴an=35+（n-3）×（-2）=41-2n
When 0 < n ≤ 20, an > 0; when n ≥ 21, an < 0
ν when Sn reaches the maximum value, n = 20
So the answer is: 20