Evaluate the following sums: sumk=120 (3k+2) and sumk=120 (k2+k).

Recall the formulas: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^{n} c = nc$.
Here, $n=20$.

$$3 \sum_{k=1}^{20} k + \sum_{k=1}^{20} 2 = 3 \left(\frac{20(20+1)}{2}\right) + (2 \times 20)$$

$$= 3 \left(\frac{20 \times 21}{2}\right) + 40$$ $$= 3 (10 \times 21) + 40$$ $$= 3(210) + 40$$ $$= 630 + 40$$ $$= 670$$

The final answer is $\boxed{670}$.

: Evaluate $\sum_{k=1}^{20} (k^2+k)$.

Recall the formulas: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.
Here, $n=20$.
For $\sum_{k=1}^{20} k^2$:

$$= \frac{20(20+1)(2 \times 20 + 1)}{6}$$ $$= \frac{20(21)(41)}{6}$$ $$= \frac{420 \times 41}{6}$$ $$= \frac{17220}{6}$$ $$= 2870$$

For $\sum_{k=1}^{20} k$:

$$= \frac{20(20+1)}{2}$$ $$= \frac{20 \times 21}{2}$$ $$= 10 \times 21$$ $$= 210$$

$$= 2870 + 210$$ $$= 3080$$

The final answer is $\boxed{3080}$.

: The 13th and 16th terms of an A.P. are 80 and 98 respectively. Find the first term, the common difference and the last term if there are 20 terms.

Given $a_{13} = 80$:

$$a + (13-1)d = 80$$ $$a + 12d = 80 \quad (1)$$

Given $a_{16} = 98$:

$$a + (16-1)d = 98$$ $$a + 15d = 98 \quad (2)$$

Subtract equation (1) from equation (2):

$$(a + 15d) - (a + 12d) = 98 - 80$$ $$3d = 18$$ $$d = \frac{18}{3}$$ $$d = 6$$

The common difference is $\boxed{6}$.

$$a + 12(6) = 80$$ $$a + 72 = 80$$ $$a = 80 - 72$$ $$a = 8$$

The first term is $\boxed{8}$.

Using the formula $a_n = a + (n-1)d$ with $n=20$, $a=8$, and $d=6$:

$$a_{20} = 8 + (20-1)6$$ $$a_{20} = 8 + (19)6$$ $$a_{20} = 8 + 114$$ $$a_{20} = 122$$

The last term is $\boxed{122}$.