If the arithmetic sequence {an} satisfies A1 = 1, and A1, A2 and A4 are equal proportion sequence, find an

If the arithmetic sequence {an} satisfies A1 = 1, and A1, A2 and A4 are equal proportion sequence, find an

Let an = 1 + D (n-1)
a1*a4=a2*a2
So 1 * (1 + 3D) = (1 + D) (1 + D)
Solve the above equation to get d = 0 or 1 (0 rounded off)
So d = 1
an=n

In the arithmetic sequence {an}, A1, A3 and A4 are equal ratio sequence, and the common ratio q?
My homework depends on you!

A2 = a1 + D, A4 = a1 + 3D, (where D is the tolerance), ∵ A1, A2, A4 these three constitute an equal ratio sequence, ∵ the square of (a1 + D) = A1 × (a1 + 3D), there is the square of D + 2 × A1 × D = 3 × A1 × D, then A1 = D, A1 = D, A2 = 2D, A4 = 4D, it is easy to know that the common ratio q is 2

On the inverse proof of summation formula of arithmetic sequence
We know Sn = (n (a1 + an)) / 2, and prove that an is an arithmetic sequence

S (n-1) = (n-1) (a1 + a (n-1)) / 2A (n) = s (n) - S (n-1) = A1 / 2 + Na (n) / 2 - (n-1) a (n-1) / 2, so (n-2) a (n) + A1 = (n-1) a (n-1) (n-1) [a (n) - A (n-1)] = a (n) - A (1) let B (n) = a (n) - A (n-1), then (n-1) B (n) = s (b (n)) (n-2) B (n-1) = s (b (n-1)) minus (n-1) B (n) - (

In the arithmetic sequence 40, 37, 34 If the first negative number in is marked as AK, then k = ()
A. 14B. 13C. 15D. 12

In the arithmetic sequence 40, 37, 34 Where, a 1 = 40, d = 37-40 = - 3, an = 40 + (n-1) × (- 3) = 43-3n, from an = 43-3n ≤ 0, n ≥ 1413, ∵ a 14 = 43-3 × 14 = 1, a 15 = 43-3 × 15 = - 2, k = 15

Formula of the number of items in arithmetic sequence
To answer today,

General term formula:
An=A1+(n-1)d
An=Am+(n-m)d
The sum of the first n terms of the arithmetic sequence:
Sn=[n(A1+An)]/2
Sn=nA1+[n(n-1)d]/2

The formula for finding the number of items in arithmetic sequence: number of items = (last first item) △ tolerance + 1, for example

5 10 15 20...255 (255-5)÷5+1=51

When the arithmetic sequence knows the sum, the number of terms, the tolerance, the first term and the last term, how to find them respectively?

The first item is A1 and the tolerance is d
The nth term is an = a1 + (n-1) d

On the sum of arithmetic series of all formulas: the number of terms, tolerance, the end of the term, etc. The more complete the better! Thank you!

Sum:
Known first term and tolerance: SN = Na1 + n (n-1) d / 2
The first and last terms are known: SN = (a1 + an) n / 2
Known final term and tolerance: SN = nan-n (n-1) d / 2
The first and last items are known
General formula: an = a1 + (n-1) d
If the m-th term is known, find the k-th term
k> When m, AK = am + (K-M) d
k

How is the formula of calculating the number of terms, tolerance, prime and sum derived from the general term formula

It is known that A1, A2 and A3 are arithmetic sequences
Number of items: an = a1 + (n-1) d
Add a formula:
an=am+(n-m)d
Tolerance: D = a2-a1
Prime Minister: A1
The formula of sum is as follows
Sn=n(a1+an)/2
Sn=na1+n(n-1)d/2
According to the formula, the known quantity is replaced and the unknown quantity is calculated

If the tolerance of an arithmetic sequence is 6, the last term is 109, and the number of terms is 18, what is the first term of the arithmetic sequence?

an=a1+(n-1)d
∴a1=an-(n-1)d=109-(18-1)×6=7
A: the first term of this arithmetic sequence is 7