Your solution to the first problem is correct but the way you have written part of it is strange. You have . What you mean, I thing, is that and then . Then you have and then replace a with 4. That you mean is so that a_1= 2.

In fact, you could have gotten that more quickly from your first equation, so that . Since d= 2, that is so .

However, you did get both d and correctly so that [tex]a_{15}= a_0+ d(15-1)= 2+ 2(15)= 30, as you say.

In (2) and (3)

The only difference between (2) and (3) is that (3) says "between 21 and 67,inclusive" so I assume that in (2), which only says "between 21 and 67" we are not to include 21 and 67 themselves. Of course, that mean that we are starting at 23 and ending at 65. A useful thing to know about arithmetic sequence is that the average ofallthe numbers in the sequence is the same as the average of the first and last numbers. The average of 23 and 65 is (23+ 65)/2= 88/2= 44. There are 65- 22= 43 numbers between 23 and 65 (inclusive) and 22 of them are odd (there is one more odd than even because both starting and ending values are odd). The sum is the average value times the number of values. You seem to have used, rather than my "average of first and last number", which is the average of all numbers down to 1.

And, of course, for (3), where you are including 21 and 67, you can just add 21+ 67= 88 to the answer for (2).