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In the arithmetic sequence {an}, if A3 + A9 + A15 + A17 = 8, then a11=______ ．
∵ sequence {an} is an arithmetic sequence. Let its tolerance be D, ∵ A3 + A9 + A15 + A17 = 8, ∵ (a1 + 2D) + (a1 + 8D) + (a1 + 14d) + (a1 + 16d) = 8, that is, 4A1 + 40d = 8, ∵ a1 + 10d = 2. That is, a11 = 2
The problem of arithmetic sequence
1. Find the value of N in the arithmetic sequence {an} a1 + A2 + a3 = 15, an + an-1 + An-2 = 78, Sn = 155
2. Given the first n terms of the sequence {an} and Sn = 3 + 2 ^ n (the nth power of 2, because I can't play), find an
A1 + A2 + a3 = 15, A2 = 5; an + an-1 + An-2 = 78, an-1 = 26;
Sn = (a1 + an) * n / 2 = (A2 + an-1) * n / 2 is substituted into the solution to get n = 10
S1 = 5 gives A1 = 5
When n > 1, an = Sn - sn-1 is substituted into an = 2 ^ (n-1)
So n = 1, an = 5; n > 1, an = 2 ^ (n-1)
Let {an} be an arithmetic sequence with tolerance D, D ≠ 0, the first 10 terms and S10 = 110, and A1, A2, A4 are equal proportion sequence. If B1 = A1, B (n + 1) = BN + an (n ∈ n +), find the general term formula of the sequence {BN}. (d = 2, an = 2n)
Hope to have the God of our school! Best before 11 o'clock~
An = 2n, then: A1 = 2, so, B1 = 2B (n + 1) = BN + an, that is: B (n + 1) = BN + 2n, then: BN = B (n-1) + 2 (n-1) B (n-1) = B (n-2) + 2 (n-2). B2 = B1 + 2 add up: sn-s1 = s (n-1) + 2 (n-1) n / 2, that is: SN-S (n-1) = B1 + n (n-1), that is: BN = B1 + n (n-1) substitute B1 = 2 into: BN = n & # - n + 2
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
From the properties and summation formula of arithmetic sequence, we can get: anbn = 2an2bn = a1 + A2N − 1B1 + b2n − 1 = (2n − 1) (a1 + A2N − 1) 2 (2n − 1) (B1 + b2n − 1) 2 = A2N − 1b2n − 1 = 7 (2n − 1) + 45 (2n − 1) + 3 = 7 + 12n + 1. We know that anbn is an integer when n = 1, 2, 3, 5, 11
Given the vector a = (- 2, - 1), B = (λ, 1), λ∈ R. (I) when λ = 3, find a · B and | a + B |; (II) if the cosine of the angle between a and B is positive, the value range of λ
(I) when λ = 3, B = (3,1), a = (- 2, - 1); ∧ a · B = - 6-1 = - 7, a + B = (1,0), | a + B | = 1; (II) the cosine of the angle between a and B is positive; ∧ a · B = - 2, λ - 1 & gt; 0; ∧ λ & lt; - 12; ∧ the value range of λ is (- ∞, - 12)
The equation of a line passing through point (2,3) with equal intercept on x-axis and y-axis______ ．
Let a line be a on the x-axis and B on the y-axis. ① when a = b = 0, the line passes through points (2, 3) and (0, 0), and the equation is YX = 32, that is, 3x-2y = 0. ② when a = B ≠ 0, the equation is XA + Ya = 1. Substituting point (2, 3), 2A + 3A = 1, a = 5, and the equation is x + Y-5 = 0. So the answer is: x + Y-5 = 0, or 3x-2y = 0
Two questions,
1: Given that the line L1 passes through a (1,1) and B (3,2), the equation of line L2 is 2x-4y-3 = 0
1. Find the position relationship between line L1 and L2, and explain the reason
2: Judge the position relation of line L1 and L2, and explain the reason
(2) It is known that the circle C: x2 + y2-2y-4 = 0 (x2y2 is x square, y Square) and the straight line L: mx-y + 1-m = 0
1. Judge the position relationship between line L and circle C
2. If the line L and circle C intersect at two different points a and B, and the absolute value of AB = 3 root sign 2, find the equation of line L
It's OK to answer only one question, which one will answer which one
1 the slope of L1 is 0.5 by two-point method, and because the slope of L2 is 0.5, the distance between two parallel lines is calculated by the formula d = root sign a ^ 2 + B ^ 2, and d = 0.5 root sign 17
2 the distance from the center of circle (0.1) to the straight line d = √ 1 / (1 + 1 / M & sup2;) is obtained by the equation of circle C, so D is less than 1, so D is less than the radius of circle C, r = √ 5, so l intersects with circle C
It is known that | a | = 3, | B | = 2, the angle between a and B is 60 °, C = 3A + 5b, d = ma + 3b and C ⊥ D
Because, C ⊥ D
So CXD = 0
（3a+5b）X（ma+3b）=0
3m|a||a|+9ab+15|b||b|+5mab=0
27m+(9+5m)|a||b|cos60'+60=0
27m+(9+5m)X3+60=0
27m+27+15m+60=0
42m=-87
m=-87/42
If f (x) = MX + n has a zero point of 2, then G (x) = NX & sup2; - MX has a zero point of 2
∵ f (2) = 0. (function f (x) = MX + n has a zero point of 2)
That is 2m + n = 0
∴n=-2m
∴g(x)=-2mx²-mx=-mx(2x+1)
Then - MX = 0 or 2x + 1 = 0
When x = 0 or - 1 / 2, G (x) = 0
The zeros of the function g (x) = NX & sup2; - MX are 0 and - 1 / 2
The solution to the maximum value of mathematical exponential function in grade one of senior high school
Given - 1 "X" 2, find the maximum and minimum of function f (x) = 3 + 2x3x + 1 (above) - 9
PS: I just don't know what to do after 3x (above) = t
Substitution is to change a complex formula into a simple unknown
In this problem, change x times of 3 into y, and then substitute it into the function
f(x)=2(y+1)-6=2y-4
∵-1≤x≤2∴1/3≤y≤9
∴f(x)min=2*1/3-4
f(x)max=2*9-4
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