I derived a new formula related to arithmetic sequences, I think!

First of all, I am a 12th grader so I don't know how to write research notes. So please forgive me if my writing is not so impressive!

I don't know what to do to tell the world about whatever I found. So, may be it's stupid, but I am posting my formula here in Stack Exchange.

Recently I was trying to figure out how to find the nth term in a series, or sequence, where the differences ($d$) between the elements of the series are not constant, but if $d$ for each $A_n - A_{(n-1)}$ is written in a sequence, it gives an AP.

For example, the sequence:

$A = 7, 14, 28, 49, ... n$

is not in AP

But taking the differences for all $A_n - A_{(n - 1)}$, where $n > 1$, we get an AP:

$$A' = 7, 14, 21...$$

The above AP has a common difference of 7.

In order to find $A_n$, We can use the below formula:

$$A_n = \left(\frac {n - 1}{2} \{2a' + (n - 2) d' \}\right) + a_1$$

Where,

$a_1$ is the first element in the sequence

$$a' = a_2 - a_1$$

$$d' = a_3 - (a' + a_2)$$

I can't describe here how I get to that point.

Here is the application of the formula in the following sequence:

$$A = 7, 14, 28, 49, ... n$$

In sequence, let $n = 3$ and we have to find $A_n$

$$a_1 = 7$$

$$a' = 14 - 7 = 7$$

$$d' = 28 - (14 + 7) = 7$$

By formula:

$$A_n = \frac {n - 1}{2} \{2a' + (n - 2) d' \} + a_1$$

$$A_3 = \frac {3 - 1}{2} \{2 \times 7 + (3 - 2) 7 \} + 7$$

$$A_3 = \frac {2}{2} \{14 + 7 \} + 7 = 28$$

Notes:

1) I am still testing it.

2) There are other methods exist to find $A_n$ in these type of sequences, may be they are easier, I don't know.

3) I am not sure if I am the first person to derive this formula, if I am not, then I apologize.

Let me know, if someone has found something wrong with this.

$$a(1)^2+b(1)+c = 7 \implies a+b+c = 7\\ a(2)^2+b(2)+c = 14 \implies 4a+2b+c=14\\ a(3)^2+b(3)+c=28 \implies 9a+3b+c = 28$$
Solving this, we get the quadratic $\frac{7}{2}x^2 -\frac{7}{2}x + 7$ which includes all of your points (specifically integers greater than or equal to $1$)
You can take this farther: Let $A^{(1)}_n=A^{(0)}_{n+1}-A^{(0)}_n$ and $A^{(2)}=A^{(1)}_{n+1}-A^{(1)}_n$ and $A^{(3)}_n=A^{(2)}_{n+1}-A^{(2)}_n$ and so on. Then $A^{(j)}_n$ is a non-zero constant if and only if $A^{(0)}_n$ is a polynomial function of $n,$ with degree $j.$ You can prove this by induction on $j.$