Find the sums of the following A.P.s as far as the terms indicate.

$$
a = 5
$$
$$
d = 10 - 5 = 5
$$
$$
n = 12
$$

$$
S_{12} = \frac{12}{2}(2(5) + (12-1)5)
$$
$$
S_{12} = 6(10 + (11)5)
$$
$$
S_{12} = 6(10 + 55)
$$
$$
S_{12} = 6(65)
$$
$$
S_{12} = 390
$$
The sum of the first 12 terms is $\boxed{\text{390}}$.

$$
a = 1\frac{1}{4} = \frac{5}{4}
$$
$$
d = 1 - \frac{5}{4} = -\frac{1}{4}
$$
$$
n = n
$$

$$
S_n = \frac{n}{2}\left(2\left(\frac{5}{4}\right) + (n-1)\left(-\frac{1}{4}\right)\right)
$$
$$
S_n = \frac{n}{2}\left(\frac{10}{4} - \frac{n-1}{4}\right)
$$
$$
S_n = \frac{n}{2}\left(\frac{10 - (n-1)}{4}\right)
$$
$$
S_n = \frac{n}{2}\left(\frac{10 - n + 1}{4}\right)
$$
$$
S_n = \frac{n}{2}\left(\frac{11 - n}{4}\right)
$$
$$
S_n = \frac{n(11 - n)}{8}
$$
The sum of the first $n$ terms is $\boxed{\frac{n(11 - n)}{8}}$.

$$
a = 1
$$
$$
d = 2 - 1 = 1
$$
$$
n = 50
$$

$$
S_{50} = \frac{50}{2}(2(1) + (50-1)1)
$$
$$
S_{50} = 25(2 + 49)
$$
$$
S_{50} = 25(51)
$$
$$
S_{50} = 1275
$$
The sum of the first 50 terms is $\boxed{\text{1275}}$.

$$
a = 2
$$
$$
d = 4 - 2 = 2
$$
$$
T_n = 46
$$

$$
46 = 2 + (n-1)2
$$
$$
46 - 2 = (n-1)2
$$
$$
44 = 2(n-1)
$$
$$
\frac{44}{2} = n-1
$$
$$
22 = n-1
$$
$$
n = 22 + 1
$$
$$
n = 23
$$
The number of terms is $\boxed{\text{23}}$.

$$
a = 2.7
$$
$$
d = 3.2 - 2.7 = 0.5
$$
$$
T_n = 17.7
$$

$$
17.7 = 2.7 + (n-1)0.5
$$
$$
17.7 - 2.7 = (n-1)0.5
$$
$$
15 = (n-1)0.5
$$
$$
\frac{15}{0.5} = n-1
$$
$$
30 = n-1
$$
$$
n = 30 + 1
$$
$$
n = 31
$$
The number of terms is $\boxed{\text{31}}$.

First term ($a$) = 1  
Common difference ($d$) = 1  
Number of terms ($n$) = $n$  
Last term ($L$) = $n$