You posted a lot of problems here. In the future please only post one problem at a time. Thank you.
I'll do the first two problems to get you started. Hopefully it will help you finish off the rest of the questions.
==========================================
Problem 1
{18, a, b, -3} is an arithmetic sequence or arithmetic progression (AP).
This means we have some number d added on to each term to get the next term.
first term = 18
second term = first term + d = 18+d = a
third term = second term + d = (18+d)+d = 18+2d = b
fourth term = third term + d = (18+2d)+d = 18+3d = -3
----
Let's solve that last equation for d
18+3d = -3
18+3d-18 = -3-18
3d = -21
3d/3 = -21/3
d = -7
----
The value d = -7 tells us to add -7 to each term to get the next term. In other words, we subtract 7 from each term to get the next term
first term = 18
second term = first term + d = 18+d = 18+(-7) = 18-7 = 11
third term = second term + d = 11+d = 11+(-7) = 11-7 = 4
fourth term = third term + d = 4+d = 4+(-7) = 4-7 = -3
----
We see that a = 11 and b = 4 are the second and third terms respectively.
Therefore, a+b = 11+4 = 15
-------------
==========================================
Problem 2
A multiple of 4 is in the form 4*n for some integer n, ie n is a whole number.
We want to know which values of 4*n are between 10 and 250.
----
Divide both 10 and 250 by 4 to get the following
10/4 = 2.5
250/4 = 62.5
If n = 2, then 4*n = 4*2 = 8 is not between 10 and 250; however n = 3 will make 4*n = 4*3 = 12 to be between 10 and 250. We see that n = 3 is the smallest possible allowed value.
If n = 62, then 4*n = 4*62 = 248 is between 10 and 250; while n = 63 will make 4*n too big because 4*63 = 252. The largest n can get is n = 62
----
The question posed in question 2 is equivalent to asking the following: "How many values are in the set {3, 4, 5, ..., 60, 61, 62}?"
You could count all of the values in the set, but that exercise is very tedious busywork. There's a much faster way. First lets consider the set below
{a, a+1, a+2, ..., b-2, b-1, b}
where a,b are integers. Basically this set starts at 'a', counts up until we get to 'b'. The handy formula
c = b-a+1
will provide the exact count of values in the set {a, a+1, a+2, ..., b-2, b-1, b}
----
In this case, a = 3 and b = 62, making
c = b-a+1
c = 62-3+1
c = 60
There are 60 values in the set {3, 4, 5, ..., 60, 61, 62}
There are 60 multiples of four that are between 10 and 250.
-------------
