How to make 3 squares with 24 matchsticks?

How to make 3 squares with 24 matchsticks?

Use two matches on each side, not one side

How many figures can five matches make
What's the matter? Say, say

Four

The sum problem of arithmetic sequence in senior high school mathematics
1. There are two arithmetic sequences 2,6,10, ···, 190 and 2,8,14, ···, 200. A new sequence is composed of the common terms of the two arithmetic sequences from small to large, and the sum of each item of the new sequence is obtained
2. It is known that the sequence (an) is an arithmetic sequence, and Sn is the sum of its first n terms. It is proved that S6, s12-s6, s18-s12 are also arithmetic sequences

1. The tolerance of the first arithmetic sequence is 4, the general term is 4n-2, and N is from 1 to 48. The tolerance of the second arithmetic sequence is 6, the general term is 6m-4, and M is from 1 to 34
The common term is 4n-2 = 6m-4, M = (2n + 1) / 3. From 1 ≤ n ≤ 48, 1 ≤ (2n + 1) / 3 ≤ 97 / 3, so m is between 1 and 32. At the same time, because 2n + 1 is odd, M can only take odd numbers: 1, 3, 5,..., 31
Therefore, the two sequences have 16 common terms, and these numbers also form an arithmetic sequence. The tolerance is 12, and the first term is 2. Therefore, the sum of each term of the new sequence is 16 × 2 + 16 * 15 / 2 × 12 = 1472
2. Suppose the first term is a and the tolerance is D, then S6 = 6A + 15d, S12 = 12a + 66d, S18 = 18a + 153d. S12-s6 = 6A + 51D, s18-s12 = 6A + 87d
(s12-s6) - S6 = 36d, (s18-s12) - (s12-s6) = 36d, so S6, s12-s6, s18-s12 are also arithmetic sequences

The first n terms and Sn of the arithmetic sequence {an} satisfy S20 = S40
A. S30 is the maximum of Sn. B. S30 is the minimum of Sn. C. S30 = 0d. S60 = 0

Let the tolerance of arithmetic sequence {an} be d. ① if d = 0, a and B can be excluded; ② if D ≠ 0, Sn = pN2 + QN (P ≠ 0), ∵ S20 = S40, ∵ 400p + 20q = 1600p + 40Q, q = - 60p, ∵ S60 = 3600p-3600p = 0; so D

How to calculate the sum of 2 + 3 + 4 +. + n with the sum formula of arithmetic sequence

If d = 1, A1 = 2, then Sn = 2n 2 (n-1) 2 / 2

Given that the sum of the first n terms of the arithmetic sequence {an} is Sn and satisfies s33-s22 = 1, then the tolerance of the sequence {an} is ()
A. 12B. 1C. 2D. 3

S3 = a1 + A2 + a3 = 3A1 + 3D, S2 = a1 + A2 = 2A1 + D, ∩ s33-s22 = D2 = 1 ∩ d = 2, so select C

Let {an} be an arithmetic sequence with non-zero tolerance, Sn be the sum of the first n terms of the sequence, S3 & # 178; = 9s2, S4 = 4s2, find the general term formula of the sequence {an}
∵ arithmetic sequence
∴S3=a1+a2+a3=3a2
S3²=9a2²=9S2
S4-S2-S2=S3~4-S2=2S2=4d
a2²=S2=a1+a2=2a1+d （1）
d=2a1 a1=d/2
Bring in (1)
d²/4+d+d²=d/2+d
Wait until d = 2 / 5
Then A1 = 1 / 5
an=a1+(n-1)d=1/5+2/5（n-1）
=(2n-1)/5
The answer is 4 / 9 (2n-1). But I don't know where the miscalculation is!

Hello, the error is (1) d & # 178 / 4 + D + D & # 178; = D / 2 + D should be: (A2) ^ 2 = 2A1 + D, that is: (a1 + D) ^ 2 = 2A1 + D (A1) ^ 2 + 2a1d + D ^ 2 = 2A1 + D. after substituting, we get: D ^ 2 / 4 + 2 * D / 2 * D + D ^ 2 = 2 * D / 2 + D. after sorting, we get: 9 / 4 * d ^ 2 = 2dd = 8 / 9, A1 = 4 / 9

If SN is the sum of the first n terms of the arithmetic sequence, and s8-s4 = 12, then the value of S12 is?

s8 - s4 =a5 + a6 + a7 + a8
a5 + a6 + a7 + a8 = 12
4a1 + 22d = 12
2a1 + 11 d = 6
a1 + (a1 + 11 d ) = 6
a1 + a12 = 6
s12 = 12 (a1 + a12 ) / 2 =12 X 6 /2 =36

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, S4 = 14, s10-s7 = 30, then S9=______ ．

∵ SN is the sum of the first n terms of the arithmetic sequence {an}, s10-s7 = 30, a8 + A9 + A10 = 30 ∵ A9 = 10, ① S4 = a1 + A2 + a3 + A4 = 2 (A2 + a3) = 2 (2A1 + 3D) = 4A1 + 6D = 14 ∵ 2A1 + 3D = 7, ② A1 = 2, d = 1 ∵ S9 = 12 × 9 × (a1 + A9) = 54, so the answer is: 54

Let Sn be the sum of the first n terms of {an} of the arithmetic sequence, and it is known that the median terms of S3 and S4 are S5, and the median terms of S3 and S4 are 1?

S5^2 = S3*S4
(S3+S4)/2 = 1=>S3+S4 = 2
S3 = a1+a1+d + a1 + 2d = 3a1 + 3d
S4 = 4a1 + 7d
S5 = 5a1 + 12d
S3 + S4 = 7a1 + 10 d = 2 => d = 1/5 - 7/10 a1
(3a1+3d)(4a1+7d) = (5a1+12d)^2
you can solve a1 and d with two equations