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If D is the midpoint of AB, then the length of DC is 10 cm, right
Right = 2 + 2 × 4 = 10cm
Given the line segment AB = 2cm, extend AB to C so that BC = 2Ab. If D is the midpoint of AB, then the length of line segment DC is______ ．
As shown in the figure: ∵ line AB = 2cm, BC = 2Ab, ∵ BC = 4cm, ∵ D is the midpoint of AB, ∵ BD = 1cm, ∵ DC = BD + BC = 1 + 4 = 5cm
As shown in the figure, given that point C is the midpoint of line AB, point D is the midpoint of BC, ab = 10cm, find the length of AD
∵ C point is the midpoint of AB, D point is the midpoint of BC, ab = 10cm, ∵ AC = CB = 12ab = 5cm, CD = 12bc = 2.5cm, ∵ ad = AC + CD = 5 + 2.5 = 7.5cm
It is known that the first term A1 of the sequence {an} is 5, and the sum of the first n terms is SN. If Sn + 1 = 2Sn + N + 5 (n ∈ n *), then the sequence {an + 1} is an equal ratio sequence. (1) write the inverse proposition of the proposition; (2) prove that the original proposition is a true proposition
(1) ∵ the original proposition is the first term A1 = 5 of the sequence {an}, and the sum of the first n terms is SN. If Sn + 1 = 2Sn + N + 5 (n ∈ n *), then the sequence {an + 1} is an equal ratio sequence; its inverse proposition is the first term A1 = 5 of the sequence {an}, and the sum of the first n terms is SN. If the sequence {an + 1} is an equal ratio sequence, then Sn + 1 = 2Sn + 5 (n ∈ n *); (2) it is proved that in the sequence {an}, A1 = 5, the sum of the first n terms is Sn, and Sn + 1 = 2Sn + 5 (n ∈ n *) The original proposition is true, that is, an + 1 = 2An + 1, an + 1 + 1 = 2An + 2, an + 1 + 1An + 1 = 2; the sequence {an + 1} is an equal ratio sequence with the common ratio q = 2 and the first term a1 + 1 = 5 + 1 = 6
The sum of the first n terms of an is Sn, A1 = 1; in BN, B1 = 1. If A3 + S3 equals 14, b2s2 = 12. Find an and BN. 2: let CN = an + 2bn (n belongs to N +), find the first n terms and TN of CN
1)an=1+(n-1)d,bn=q^(n-1)
a3+s3=2+4d+1+1+d=4+5d=14,d=2,an=2n-1
b2s2=q(1+1+d)=4q=12,q=3,bn=3^(n-1)
2)cn=2n-1+2*3^(n-1)
Tn =(1+2n-1)n/2+2+2*3+...+2*3^(n-1)
3TN=3n^2 +2*3+...+2*3^(n-1)+2*3^n
2Tn=2n^2+2*3^n-2
Tn=n^2+3^n-1
All the items of the arithmetic sequence {an} are positive, A1 = 1, the sum of the first n terms is Sn, the sequence {BN} is equal ratio sequence, B1 = 1, and b2s2 = 6, B2 + S3 = 8 (1) the general formula of autumn sequence {an} and {BN} (2) finding 1 / S1 + 1 / S2 + L + 1 / Sn
The second question is 1 / S1 + 1 / S2 + +1/SN
(1) Suppose the tolerance of the arithmetic sequence {an} is D, and the common ratio of the arithmetic sequence {BN} is Q. then according to the meaning of b2s2 = 6, B2 + S3 = 8, q (1 + 1 + D) = 6q + (1 + 1 + D + 1 + D + D) = 8, q (2 + D) = 6. ① Q + 3D = 5
Let the sum of N before {an} be Sn, a3-a2 = 1, S1, S3, S4 be an arithmetic sequence, and find the general formula of {an}!
First of all, it is known that the equal ratio sequence with common ratio q = 1 does not satisfy the meaning of the problem! So suppose that Q ≠ 1, then suppose that the first term is a, the common ratio is Q, and the general term formula is an = AQ ^ (n-1)
In the equal ratio sequence {an}, it is known that S3 = 9 / 2, A3 = 3 / 2, to find A1 and Q
S3=a1+a2+a3=a1+a1q+a1q^2=a1(1+q+q^2)=9/2 (1)
a3=a1q^2=3/2 (2)
(1) / (2) 2q ^ 2-q-1 = 0 (Q-1) (Q + 1 / 2) = 0, q = 1 or q = - 1 / 2
When q = 1, A1 = 3 / 2
When q = - 1 / 2, A1 = 6
In the equal ratio sequence {an}, A3 = 1 1 / 2, S3 = 4 1 / 2, find A1 and Q
When A3 = A1 * q ^ 2 = 1,1 / 2 = 3 / 2A1 = 3 / (2q ^ 2) S3 = a1 + A1 * q + 3 / 2 = 4,1 / 2 = 9 / 2A1 + a1q = 33 / (2q ^ 2) + 3 / (2q) = 33 + 3Q = 6q ^ 22q ^ 2-q-1 = 0 (2q + 1) (Q-1) = 0q = - 1 / 2 or a = 1q = - 1 / 2, A1 = 3 / (1 / 2) = 6q = 1, A1 = 3 / 2
If A1 = 2 and S3 = 6 are known in the equal ratio sequence an, find A3 and Q
A1 = 2, so A2 = 2q, A3 = 2q ^ 2, so 22q 2q ^ 2 = 6, q = 1 or - 2 of the solution, so A3 = 2 or 8
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