The number pattern below shows the sum of consecutive odd numbers. Complete lines 5 and 6, find the last number in line 10, and the sum in line 12.

The pattern shows the sum of consecutive odd numbers starting from 1.
Line $n$ sums the first $n$ odd numbers.
The sum for Line $n$ is $n^2$.
The last number in Line $n$ is the $n$-th odd number, given by $2n-1$.

Line 5 will sum the first 5 odd numbers. The 5th odd number is $2(5)-1 = 9$.
The sum for Line 5 is $5^2 = 25$.

Line 6 will sum the first 6 odd numbers. The 6th odd number is $2(6)-1 = 11$.
The sum for Line 6 is $6^2 = 36$.

The completed lines are:
Line 5: $1+3+5+7+9 = 25 = 5 \times 5$
Line 6: $1+3+5+7+9+11 = 36 = 6 \times 6$

The last number in Line $n$ is the $n$-th odd number, which is given by the formula $2n-1$.

For Line 10, $n=10$.
Last number $= 2(10)-1 = 20-1 = 19$.
The last number in Line 10 would be $\boxed{\text{19}}$.

The sum of the numbers in Line $n$ is $n^2$.

For Line 12, $n=12$.
Sum $= 12^2 = 144$.
All the numbers in Line 12 would add up to $\boxed{\text{144}}$.

From the pattern, the sum of the first $n$ odd numbers is $n^2$.

To add up the first 20 odd numbers, you simply square the number of odd numbers, which is 20.
So, the sum is $20^2 = 400$.
The quick way to add up the first 20 odd numbers is to square the number 20, which gives $\boxed{\text{400}}$.