In the equal ratio sequence {an}, if a1 + A2 + a3 = 7, a1a2a3 = 8, and the common ratio is greater than 1, find the first sequence N and Sn of the sequence
a1a3=a2^2
a1a2a3=8 a2=2
a1+a2+a3=7
a1+a3=5
a1*a3=4
a1=1 a3=4 q=2
A1 = 4 A3 = 1 Q = 1 / 2 (rounding)
Sn=a1(1-q^n)/(1-q)
=2^n-1
RELATED INFORMATIONS
- 1. Only the first N-term sum formula of the sequence {an} is Sn = (3) ^ n - 1, the general term formula of {an} is obtained, and it is proved that {an} is an equal ratio sequence
- 2. Given that the sum of the first n terms of the arithmetic sequence {an} is Sn = 2n & # 178; - 10N, find the values of (1) A1 and A3; (2) find A5 + A6 + A7 + A8; (3) find its general term formula and judge whether it is arithmetic sequence
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- 4. If the tolerance of arithmetic sequence {an} d = 1 / 2, a1 + A2 + a3... + A99 = 60, calculate S100
- 5. In the arithmetic sequence {an}, d = 12, S100 = 145, then a1 + a3 + A5 + +The value of A99 is () A. 57B. 58C. 59D. 60
- 6. The tolerance of arithmetic sequence an d = 1 / 2, and S100 = 145, find a1 + a3 + A5 +. + A99 Why: M2 = A2 + A4 + A6 +... + A100 = a1 + a3 + A5 +... + A99 + (1 / 2) × 50 How does this (1 / 2) × 50 come from?
- 7. It is known that {an} is an arithmetic sequence with a tolerance of 2, and A1, A3 and A4 are proportional sequences, then the sum of the first nine terms of the sequence {an} is equal to () A. 0B. 8C. 144D. 162
- 8. Known: in the arithmetic sequence {an}, A3 + A4 = 15, a2a5 = 54, tolerance d < 0 Find the maximum value of [Sn - (an-3)] / N and the corresponding value of n Ans:n=4 In this case, ymax = 15 / 2
- 9. Known: in the arithmetic sequence {an}, A3 + A4 = 15, a2a5 = 54, tolerance d < 0, the second question 1) Finding the general term formula an of sequence {an} 1、 The arithmetic sequence is A2 + A5 = A3 + A4 = 15 a2a5=54 By Weida theorem A2, A5 are the equations X & # 178; - 15x + 54 = 0 x=6,x=9 d
- 10. If A2 = 1, A3 = 3, then S4 = () A. 12B. 10C. 8D. 6
- 11. Given that the sequence {an} satisfies A1 = 1, an + 1 = 2An + 1 (n ∈ n +), (1) let BN = an + 1, prove that the sequence {BN} is an equal ratio sequence; (2) find the expression of an
- 12. If the first n terms of positive term sequence {an} and Sn satisfy 10sn = an ^ 2 + 5An + 6, and A1, A3, A15 are equal proportion sequence, then A2010=
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- 14. Given the positive term sequence {an}, the first n terms and Sn satisfy 10sn = an2 + 5An + 6, and A1, A3, A15 are equal proportion sequence, find the general term an of the sequence {an}
- 15. Given the positive term sequence {an}, the first n terms and Sn satisfy 10sn = an2 + 5An + 6, and A1, A3, A15 are equal proportion sequence, find the general term an of the sequence {an}
- 16. Given the positive term sequence {an}, the first n terms and Sn satisfy 10sn = an2 + 5An + 6, and A1, A3, A15 are equal proportion sequence, find the general term an of the sequence {an}
- 17. In the equal ratio sequence {an}, A1 = 2, A4 = 16 are known
- 18. Given that the sequence {an} is an equal ratio sequence, a1 + a3 = 10, A4 + A6 = 5 / 4, find the value of A5 and the first n terms and S6
- 19. In the equal ratio sequence an, a1 + a3 = 10, A4 + A6 = 5 / 4, the value of common ratio q is
- 20. In the known sequence {an}, A1 = 1, an + 1 = 2An + 1, let BN = an + 1-an. (1) prove that the sequence {BN} is an equal ratio sequence; (2) let the sum of the first n terms of the sequence {Nan} be Sn, and find the minimum value of positive integer n that makes Sn + n (n + 1) 2 > 120