If a1 + A5 = 2, A3 + A7 = 6, then A5 + A9

If a1 + A5 = 2, A3 + A7 = 6, then A5 + A9


eighteen
Obviously, the equivalent ratio is 1.732



The sequence {an} is an arithmetic sequence with non-zero tolerance, and A5, a8 and A13 are three adjacent terms of the arithmetic sequence {BN}. If B2 = 5, then BN = ()
A. 5•(53)n−1B. 5•(35)n−1C. 3•(35)n−1D. 3•(53)n−1


∵ {an} is an arithmetic sequence with non-zero tolerance, and A5, A8, A13 are the adjacent three terms of the arithmetic sequence {BN}, ∵ (A5 + 3D) 2 = A5 (A5 + 8D), ∵ A5 = 92d, ∵ q = A5 + 3da5 = 152d92d = 53, ∵ B2 = 5, q = 53, ∵ B1 = b2q = 3, ∵ BN = b1qn − 1 = 3 · (53) n − 1



The sequence {an} is an arithmetic sequence with non-zero tolerance, and A5, a8 and A13 are the adjacent three terms of the arithmetic sequence {BN}. (1) find the common ratio of the arithmetic sequence {BN}


an=a1+(n-1)d
a5=a1+4d
a8=a1+7d
a13=a1+12d
a5*a13=(a8)²
(a1+4d)*(a1+12d)=(a1+7d)²
d=2a1
a5=a1+4d=9a1
a8=a1+7d=15a1
a13=a1+12d=25
Common ratio q = 15:9 = 5 / 3



Let the first term A1 = 256 of the equal ratio sequence {an}, the sum of the first n terms be Sn, and Sn, Sn + 2, Sn + 1 form the equal difference sequence. (I) find the common ratio Q (2) of {an}, and use iin to express {an}
The product of the first n terms of
Iin = A1 * A2... An, try to compare the size of ii7, II8, ii9
For the known function f (x) = x + T / X (T > 0) and point P (1,0), two tangent lines PM and PN of the curve y = f (x) are made through point P, and the tangent points are m and N respectively
(1) Let / Mn / = g (T) try to find the expression of the function g (T);
(2) Whether there is t, so that m, N and a (1,0) are collinear. If there is, find out the value of T; if not, explain the reason
『3』 Under the condition of (1), if for any positive integer n, there are always m + 1 real numbers A1, A2,..., am, am + 1 in the interval [2, N + 64 / N], such that the inequality g (A1) + G (A2) +... + G (AM)


(n + 2) = Sn + s (n + 1 (n + 1 + 1) + a (n + 2)] = Sn + Sn + a (n + 1 (n + 1) + a (n + 1 + 1) + 2 (n + 2)] (Sn + n + 1 (n + 1) + a (n + 1 + 1) 2Sn + 2 (n + 1) 2Sn + 2 (n + 1) 2 (n + 1 + 1) + 2 [Sn + a (n + 1 + 1) + a (n + 1 (n + 1) + a (n + 2)] (Sn + 1 (n + 1 (n + 1) 2Sn + 1) 2Sn + 2Sn + 1 (n + 1) 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2Sn + 2 (n + 2 it's not easy