Formula for sum to $n$ terms in an AP

Question

I have always been taught that the sum to $n$ terms of an AP is given by:

$$S_n=\frac n2 [2a+(n-1)d]$$

(where $a$ is first term and $d$ is common difference)

From this formula, it is very easy to derive

$$S_n=n^2(\frac d2)+n(a-\frac d2)$$

This is more helpful in solving some questions such as

1. If $S_n=3n^2+5n$, find $a_7$.
2. If the ratio of sum to $n$ terms of two APs is $\frac{n+6}{n+5}$, find ratio of third terms.

Normal method

1.

$a=S_1=3(1)^2+5(1)=8$

$S_2=3(2)^2+5(2)=22$

$a_2=S_2-S_1=14$

$d=a_2-a_1=6$

Now we find $a_7$ using $a_7=a+6d$

2.

This is also done by a lengthy method that I don't exactly know.

Methods using second formula (which I derived)

1.

By comparison, $\frac d2=3$ and $a-\frac d2=5$.

So $d=6$. Putting this in this other equation gives $a=8$

2.

$$\frac{S_n}{S_n'}=\frac{n^2(\frac d2)+n(a-\frac d2)}{n^2(\frac > {d'}2)+n(a'-\frac{d'}2)}=\frac{n(\frac d2)+(a-\frac d2)}{n(\frac > {d'}2)+(a'-\frac{d'}2)}$$

By comparison, $\frac d2=1$, $a-\frac d2=6$, $\frac {d'}2=1$ and $a-\frac {d'}2=5$

Hence $d=2$, $a=7$, $d'=2$ and $a'=6$

Using this info, we find the ratio.

Does my method work? And is there any reason why this formula is not taught instead of the usual one?