- #1
Astronomer107
- 31
- 0
I've been working on these problems for a long time trying to prove to myself that I'm not stupid (though I am convinced I am because I don't feel like "IB material" at the moment), but this will be impossible since I have other things to do. So, my two questions are these:
1. An arithmetic sequence has a common difference of 2 and a sum of 120. The first term is numerically equal to the number of terms. FInd all possible values of hte first.
I know that the answer is 8 because it is a multiple choice question and through elimination, the others aren't possible, but I don't understand how to prove this mathematically.
2. Determine Sn of the series 4 + 10 + 16 + 22+...+ (6n - 2).
I tried to solve this one, but when I did, the n's canceled out, making it "all real numbers" and since (6n - 2) is the last term, the series isn't infinite.
PLEASE HELP ME! Thanks!
1. An arithmetic sequence has a common difference of 2 and a sum of 120. The first term is numerically equal to the number of terms. FInd all possible values of hte first.
I know that the answer is 8 because it is a multiple choice question and through elimination, the others aren't possible, but I don't understand how to prove this mathematically.
2. Determine Sn of the series 4 + 10 + 16 + 22+...+ (6n - 2).
I tried to solve this one, but when I did, the n's canceled out, making it "all real numbers" and since (6n - 2) is the last term, the series isn't infinite.
PLEASE HELP ME! Thanks!