As shown in the figure, in △ MPN, MP = NP, ∠ MPN = 90 °, NQ ⊥ PQ, MS ⊥ PQ, the perpendicular feet are Q and s respectively. (1) try to explain: △ PMS ≌ △ NPQ; (2) if QS = 3.5cm, NQ = 2.1cm, calculate the length of MS

As shown in the figure, in △ MPN, MP = NP, ∠ MPN = 90 °, NQ ⊥ PQ, MS ⊥ PQ, the perpendicular feet are Q and s respectively. (1) try to explain: △ PMS ≌ △ NPQ; (2) if QS = 3.5cm, NQ = 2.1cm, calculate the length of MS

(1) ∫ MPN = 90 °, NQ ⊥ PQ, MS ⊥ PQ, ∫ PSM = ≁ q = ≁ MPN = 90 °, ∫ SPM + ≌ PMS = 90 °, ∫ SPM + ≌ NPQ = 90 °, ≁ PMS = ≁ NPQ, in △ PMS and △ NPQ, ≁ PSM = ≁ Q ≁ PMS = ≌ NPQ (AAS); (2) ∵ QS = 3.5cm, NQ = 2

Simplify OP QP + MS NQ (all vectors)

Vector on + vector MS

Using vector knowledge to prove (MP + NQ) ^ 2

Let a = (m, n), B = (P, q)
A * b = | a | * | B | cos angle|
(a*b)^2

In the arithmetic sequence, the tolerance D

I don't want to
If we only know the condition A3A5 + a3a7 + a5a9 + a7a9 = 0, we have to think from here
After analysis
It can be factorized
a3（a5+a7）+a9（a5+a7）=0
（a5+a7)(a3+a9)=0
So A5 + A7 = 0 or A3 + A9 = 0
There are also 2A6 = A5 + A7 = 0 or 2A6 = A3 + A9 = 0
That is, A6 is equal to 0 anyway
So the maximum is 2 × 6-1, which is n = 11

10,8,6,4

An=10-2(n-1)
d=-2
A8=10-2*7=-4
A10=10-2*9=-8
S10=(10-8)*10/2=10

Suppose that each item of the arithmetic sequence {an} is an integer, and its tolerance D ≠ 0, A5 = 6, A2 × A10 > 0, find the value of D

Every term of {an} is an integer, so D is an integer, and D ≠ 0
a5=6 →
a2=6-3d
a10=6+5d
a2*a10=(6-3d)(6+5d)>10→
36+2d-15d^2>10→
15d^2-2d-260
When d = - 1, A2 = 9, A10 = 1, A2 × A10 > 0

If the three sides a, B and C of a straight triangle form an arithmetic sequence and the tolerance d > 0, then a / d

If the three sides of a right triangle a, B, C form an arithmetic sequence and the tolerance d > 0, then a / D, according to the meaning of the problem, let the three sides be: B-D, B, B + D. because the tolerance d > 0, so (B + D) is the hypotenuse of RT △ ABC, according to the Pythagorean theorem: (B + D) & sup2; - (B-D) & sup2; = B & sup2; = = = = = > 4bd = B & sup2; = = = = = > b / D =

Mathematical proof: sum of each k item of arithmetic sequence SK, S2K SK, s3k-s2k The tolerance is k ^ 2 times of the original tolerance
I have a poor IQ,

Certification:
Use the definition of arithmetic sequence
Let the tolerance of arithmetic sequence {an} be d
Then SK, S2K SK, s3k-s2k The general term of is BN = a (nk-k + 1) + a (nk-k + 2) +. + a (NK)
∴ b(n+1)= a(nk+1)+a(nk+2)+.+a(nk+k)
∴ b(n+1)-b(n)
=[a(nk+1)+a(nk+2)+.+a(nk+k)]-[ a(nk-k+1)+a(nk-k+2)+.+a(nk)]
=[a(nk+1)-a(nk-k+1)]+[a(nk+2)-a(nk-k+2)]+.+[a(nk+k)-a(nk)]
= kd + kd +.+ kd
There are K in total
=K & # 178; D (is a constant)
The sum of K items in the arithmetic sequence is SK, S2K SK, s3k-s2k The tolerance is k ^ 2 times of the original tolerance

The sequence {an} is an arithmetic sequence with tolerance D. It is proved by definition that the sequence {a (4n-3)} is an arithmetic sequence

The recurrence formula is an = a1 + (n-1) d
The recursive formula a (4n-3) = a1 + (n-1) * 4D is replaced by 4n-3
Then a (4n-3) is also an arithmetic sequence with a tolerance of 4D

How to prove that the tolerance of arithmetic sequence composed of 7 prime numbers is greater than or equal to 210

If the tolerance is not a multiple of 3, then one of the three consecutive numbers must be a multiple of 3; if the tolerance is not a multiple of 5, then one of the five consecutive numbers must be a multiple of 5. Similarly, the tolerance of these seven numbers must be a multiple of 2, 3, 5 and 7
Counter example, 715730745760757907