As shown in the figure, ad = 3.5cm, BC = 3cm, points D and E are the midpoint of AC and ab respectively, and the length of line EC is calculated

As shown in the figure, ad = 3.5cm, BC = 3cm, points D and E are the midpoint of AC and ab respectively, and the length of line EC is calculated

It is proved that in △ abd and △ ACD
AB = AC
BD = CD
AD = AD
∴△ABD≌△ACD（SSS）
∴∠1 =∠2
In △ Abe and △ ace
AB = AC
∠1 =∠2
AE = AE
∴△ABE≌△ACE（SAS）
∴EB = EC

As shown in the figure, ab = CD, e and F are the midpoint of BC and ad respectively. The ray CD and EF intersect at the point h, and the verification is: ∠ Bge = ∠ Che
 

Prove: connect BD, take the midpoint m of BD, connect EM FM, because e and F are the midpoint of BC and ad, so EM FM is the median line of triangle BCD and triangle abd, so me = 1 / 2cdme parallel CD, so MEF = angle chemf = 1 / 2abmf parallel AB, so MFE = angle Bge, because AB = CD, so me = MF, so MFE = angle M

As shown in the figure, B and C are the two points on the ray ad, e is the midpoint of the line AB, f is the midpoint of the line CD, ad = 18cm, BC = 5C
Find AB + CD
Finding the distance between two points of E.F

ab+cd=18-5=13cm ef=5+6.5=11.5cm

It is known that the sequence {an} is an arithmetic sequence, the partial terms of tolerance D ≠ 0 and {an} constitute the following sequence: AK1, ak2 Where K1 = 1, K2 = 5, K3 = 17, find K1 + K2 + K3 + +kn．

Let the first term of {an} be A1, ∵ AK1, ak2, ak3 into an equal ratio sequence, ∵ a1 + 4D) 2 = A1 (a1 + 16d). We obtain A1 = 2D, q = ak2ak1 = 3. ∵ AKN = a1 + (kn-1) d, AKN = A1 · 3n-1, ∵ kn = 2 · 3n-1-1. ∵ K1 + K2 + +kn=2（1+3+… +3n-1）-n=2×1−3n1−3-n=3n-n-1．

It is known that {an} is an arithmetic sequence with tolerance D ≠ 0. Some terms AK1, ak2 If K1 = 1, K2 = 5, K3 = 17, (1) find kn; (2) find K1 + 2k2 + 3K3 + +nkn．

(1) Let AK1, ak2 The common ratio of AKN is Q ∵ K1 = 1, K2 = 5, K3 = 17 ∵ A1 · A17 = a52, that is, A1 (a1 + 16d) = (a1 + 4D) 2, & nbsp; it is obtained that a1d = 2d2 ∵ D ≠ 0 ∵ A1 = 2D, q = a5a1 = 3 ∵ AKN = a1 + (KN − 1) d = (KN + 1) d, AKN = AK1 · QN − 1 = 2D × 3N − 1

In the arithmetic sequence {an}, the tolerance D ≠ 0, A2 is the median of the ratio of A1 and A4, the known sequence A1, A3, AK1, ak2 ，akn，… The general term kN of the sequence {kn} can be obtained

According to the meaning of the question, A22 = a1a4, i.e. (a1 + D) 2 = A1 (a1 + 3D) and D ≠ 0, A1 = D and A1, A3, AK1, ak2, AKN, form an equal ratio sequence, and the common ratio of the sequence is q = a3a1 = 3dd = 3, so AKN = A1 · 3N + 1 and AKN = a1 + (KN − 1) d = KNA1  kn = 3N + 1, so the general term of the sequence {kn} is kn = 3N + 1

It is known that the sequence {an} is an arithmetic sequence, the partial terms of tolerance D ≠ 0 and {an} constitute the following sequence: AK1, ak2 AKN is a sequence of equal ratio
Where K1 = 1, K2 = 6, K3 = 26
(1) find the general term formula of the sequence {kn} (2) find the sum of the first n terms of the sequence {kn}

(1) Let an = a1 + (n-1) d Formula 1, ak3 / ak2 = ak2 / AK1 = q formula 2
From the question, there are AK1 = A1, ak2 = A6, ak3 = A26,
From equation 2, there is A1 (a1 + 25d) = (a1 + 5d) ^ 2, and the solution is A1 = 5 / 3D;
In Formula 1, A1 = 5 / 3D, A6 = 20 / 3D, A26 = 80 / 3D, q = 4 is obtained from formula 2;
So AKN = 5 / 3D + (kn-1) d = (5 / 3D) * q ^ (n-1), that is, 5 / 3 + kn-1 = 5 / 3 * 4 ^ (n-1);
The solution is kn = 5 / 3 * 4 ^ (n-1) - 2 / 3
(2)Sn=5/3*[1*(4^n-1)]/(4-1) - 2/3n=5/9 * (4^n-1)-2/3n

It is proved that {2 ^ an} is an arithmetic sequence

{an} is an arithmetic sequence
So an-a (n-1) = D, D is a fixed value
2^an÷2^[a(n-1)]=2^[an-a(n-1)]=2^d
If D is the fixed value, then 2 ^ D is the fixed value
That is to say, the latter is a fixed value except the former
So {2 ^ an} is an equal ratio sequence

If M + n = P + Q, M N P Q ∈ n *, there is am + an = AP + AQ in the arithmetic sequence, what about in the arithmetic sequence?

a[m]*a[n]=a[p]*a[q]

If M + n = P + Q, am + an = AP + AQ, and vice versa? Why?
n. M, P, q are all subscripts!

It doesn't hold if a = 0