e. n^th term is: f. n^th term is: g. n^th term is: Make notes on how you solved the sequences.

The terms are $-3, 6, 15, 24$.
The difference between consecutive terms is $6 - (-3) = 9$, $15 - 6 = 9$, $24 - 15 = 9$.
The common difference is 9.

Since the common difference is 9, the $n^{th}$ term will be of the form $9n + c$.
For $n=1$, the term is $-3$. So, $9(1) + c = -3 \implies 9 + c = -3 \implies c = -12$.
Thus, the $n^{th}$ term is $9n - 12$.

Substitute $n=10$ into the $n^{th}$ term formula:
$T_{10} = 9(10) - 12 = 90 - 12 = 78$.

The $n^{th}$ term is: $\boxed{\text{9n - 12}}$
The $10^{th}$ term is: $\boxed{\text{78}}$

The terms are $13, 17, 21, 25$.
The difference between consecutive terms is $17 - 13 = 4$, $21 - 17 = 4$, $25 - 21 = 4$.
The common difference is 4.

Since the common difference is 4, the $n^{th}$ term will be of the form $4n + c$.
For $n=1$, the term is 13. So, $4(1) + c = 13 \implies 4 + c = 13 \implies c = 9$.
Thus, the $n^{th}$ term is $4n + 9$.

Substitute $n=10$ into the $n^{th}$ term formula:
$T_{10} = 4(10) + 9 = 40 + 9 = 49$.

The $n^{th}$ term is: $\boxed{\text{4n + 9}}$
The $10^{th}$ term is: $\boxed{\text{49}}$

The terms are $-6, 10, 26, 42$.
The difference between consecutive terms is $10 - (-6) = 16$, $26 - 10 = 16$, $42 - 26 = 16$.
The common difference is 16.

Since the common difference is 16, the $n^{th}$ term will be of the form $16n + c$.
For $n=1$, the term is $-6$. So, $16(1) + c = -6 \implies 16 + c = -6 \implies c = -22$.
Thus, the $n^{th}$ term is $16n - 22$.

Substitute $n=10$ into the $n^{th}$ term formula:
$T_{10} = 16(10) - 22 = 160 - 22 = 138$.

The $n^{th}$ term is: $\boxed{\text{16n - 22}}$
The $10^{th}$ term is: $\boxed{\text{138}}$

Problem solving - First sequence
The given terms are $T_3 = 13$, $T_6 = 40$, $T_9 = 85$, $T_{12} = 148$.

Let the $n^{th}$ term be $T_n = an^2 + bn + c$.
Using the given terms:
For $n=3$: $9a + 3b + c = 13$ (Eq 1)
For $n=6$: $36a + 6b + c = 40$ (Eq 2)
For $n=9$: $81a + 9b + c = 85$ (Eq 3)

Subtract (Eq 1) from (Eq 2): $27a + 3b = 27$ (Eq 4)
Subtract (Eq 2) from (Eq 3): $45a + 3b = 45$ (Eq 5)
Subtract (Eq 4) from (Eq 5): $18a = 18 \implies a = 1$.
Substitute $a=1$ into (Eq 4): $27(1) + 3b = 27 \implies 3b = 0 \implies b = 0$.
Substitute $a=1, b=0$ into (Eq 1): $9(1) + 3(0) + c = 13 \implies 9 + c = 13 \implies c = 4$.

The $n^{th}$ term is $T_n = 1n^2 + 0n + 4 = n^2 + 4$.

Substitute $n=10$ into the $n^{th}$ term formula:
$T_{10} = 10^2 + 4 = 100 + 4 = 104$.

The $n^{th}$ term is: \boxed{\text{n^2 + 4}}
The $10^{th}$ term is: $\boxed{\text{104}}$

Problem solving - Second sequence
The given terms are $T_6 = 199$, $T_{12} = 1711$, $T_{18} = 5815$. We need to find $T_{10}$.

Let the $n^{th}$ term be $T_n = an^2 + bn + c$.
Using the given terms:
For $n=6$: $36a + 6b + c = 199$ (Eq 1)
For $n=12$: $144a + 12b + c = 1711$ (Eq 2)
For $n=18$: $324a + 18b + c = 5815$ (Eq 3)

Subtract (Eq 1) from (Eq 2): $108a + 6b = 1512$ (Eq 4)
Subtract (Eq 2) from (Eq 3): $180a + 6b = 4104$ (Eq 5)
Subtract (Eq 4) from (Eq 5): $72a = 2592 \implies a = 36$.
Substitute $a=36$ into (Eq 4): $108(36) + 6b = 1512 \implies 3888 + 6b = 1512 \implies 6b = -2376 \implies b = -396$.
Substitute $a=36, b=-396$ into (Eq 1): $36(36) + 6(-396) + c = 199 \implies 1296 - 2376 + c = 199 \implies -1080 + c = 199 \implies c = 1279$.

The $n^{th}$ term is $T_n = 36n^2 - 396n + 1279$.

Substitute $n=10$ into the $n^{th}$ term formula:
$T_{10} = 36(10^2) - 396(10) + 1279$
$T_{10} = 36(100) - 3960 + 1279$
$T_{10} = 3600 - 3960 + 1279$
$T_{10} = -360 + 1279 = 919$.

The $n^{th}$ term is: \boxed{\text{36n^2 - 396n + 1279}}
The $10^{th}$ term is: $\boxed{\text{919}}$

2. Make notes on how you solved the sequences.
For arithmetic sequences (like parts a-g), I found the common difference ($d$) between consecutive terms. The $n^{th}$ term was then determined using the formula $T_n = a + (n-1)d$, where $a$ is the first term.
For quadratic sequences (like the "Problem solving" sections), I used the general form $T_n = an^2 + bn + c$. By substituting the given term positions ($n$) and their values ($T_n$), I created a system of simultaneous equations to solve for the coefficients $a, b,$ and $c$.