It is known that the fine brushwork Q of equal ratio sequence is not equal to 1, and am, an and AP are equal ratio sequence. It is proved that m, N and P are equal difference sequence
Because am, an and AP form a series of equal proportion numbers, then the middle term of equal proportion consists of:
(an)^2=am*ap
(A1 * q ^ (n-1)) ^ 2 = A1 * q ^ (m-1) * A1 * q ^ (p-1)
Then cancel A1, (Q ^ (n-1)) ^ 2 = q ^ (m-1) * q ^ (p-1)
Because Q ≠ + - 1
SO 2 (n-1) = (m-1) + (p-1)
That is, 2n = m + P
It can be explained that m, N and P are equal difference sequence
RELATED INFORMATIONS
- 1. The sequence {an} is an equal ratio sequence, m, N, P are equal difference sequence, if am = 4, an = 6, then the value of AP is equal
- 2. Three real numbers that are not equal to each other form an arithmetic sequence. If these three numbers are properly arranged, they can become an equal ratio sequence, and the sum of these three numbers is 6, then these three numbers can be calculated
- 3. Three unequal real numbers form an arithmetic sequence. After properly exchanging the positions of the three numbers, they become an arithmetic sequence. Then the common ratio of the arithmetic sequence is___ .
- 4. Three unequal real numbers form an equal difference sequence, and then form an equal ratio sequence after proper exchange of positions, so as to find the common ratio of the equal ratio sequence
- 5. There are four real numbers. The first three are in equal proportion sequence, and their product is 216. The last three are in equal difference sequence, and their sum is 12 Set up four unknowns!
- 6. There are four real numbers. The first three numbers form an equal ratio sequence, and their product is - 64. The last three numbers form an equal difference sequence, and their product is - 640 Find these four numbers, find the detailed process
- 7. It is known that the first three numbers of four positive real numbers are equal difference sequence, the last three numbers are equal ratio sequence, the sum of the first and the third is 8, and the product of the second and the fourth is 36 (II) if the first three numbers are the first three terms of the arithmetic sequence an and the last three numbers are the first three terms of the proportional sequence BN, let CN= an.bn To find the first n terms and TN of the sequence CN
- 8. 3x + 1 / 4 = 3 / 5 solution equation
- 9. The solution equation: (40-0.8x) 178; + (30-0.5x) 178; = 16 & # 178;
- 10. There are four real numbers. The product of the first three arithmetical sequences is 216, and the sum of the last three arithmetical sequences is 24
- 11. In the equal ratio sequence {an}, the sum of the first n terms is SN. If SM, SM + 2, SM + 1 are equal difference sequence, then am, am + 2, am + 1 are equal difference sequence. (1) write the inverse proposition of this proposition; (2) judge whether the inverse proposition is true? The proof is given
- 12. It is known that SM and Sn respectively represent the sum of the first m terms and the first n terms of the arithmetic sequence {an}, and smsn = m2n2, then aman=______ .
- 13. Given that the sum of the first n terms of two arithmetic sequences {an} and {BN} is an and BN respectively, and anbn = 7n + 45N + 3, then the number of positive integers n with anbn as an integer is () A. 2B. 3C. 4D. 5
- 14. Given that the sum of the first n terms of two arithmetic sequences {an} and {BN} is an and BN respectively, and anbn = 7n + 45N + 3, then the number of positive integers n with anbn as an integer is () A. 2B. 3C. 4D. 5
- 15. Given that the sum of the first n terms of two arithmetic sequences {an} and {BN} is an and BN respectively, and anbn = 7n + 45N + 3, then the number of positive integers n with anbn as an integer is () A. 2B. 3C. 4D. 5
- 16. If the sum of the first n terms of two arithmetic sequences {an} and {BN} is an and BN respectively, and an / BN = 7n + 45 / N + 3, then the number of positive integers n with an / BN as an integer is? First of all, thank you
- 17. Given that the sum of the first n terms of two arithmetic sequences {an} and {BN} is an and BN respectively, and anbn = 7n + 45N + 3, then the number of positive integers n with anbn as an integer is () A. 2B. 3C. 4D. 5
- 18. It is known that the arithmetic sequence {an} satisfies the following conditions: A3 + A4 = 16, A4 + A5 = 20, and the sum of the first n terms of {an} is SN (I) find SN (II) finding the first n terms and TN of sequence {1 / Sn} I can do the first question,
- 19. The sum of the first n terms of the arithmetic sequence {an} is Sn, and a4-a2 = 8, A3 + A5 = 26. Note TN = snn2, if there is a positive integer m, so that TN ≤ m holds for all positive integers n, then the minimum value of M is______ .
- 20. It is known that the positive term sequence {an}, {BN} satisfies the following conditions: for any positive integer n, an, BN, a (n + 1) are equal difference sequence, BN, a (n + 1), B (n + 1) are equal ratio sequence, And A1 = 10, A2 = 15 Verification: sequence (root BN) is arithmetic sequence General formula for solving sequence {an}, {BN} Let Sn = 1 / (A1) + 1 / (A2) + 1 / (A3) +. 1 / (an) for any positive integer n, the inequality 2asn