Precal Help!!?

Question

Please someone help, I have no clue on how to work these out and I would appreciate it if some one can show me how to work them out or just give me the answer and I try to figure it out. Please someone help!!!

1. For the sequence an=an-1+an-2
and if a1=2 and a2=3
its third term is
its fourth term is
its fifth term is
**here the n-1 and n-2 are subscript**

2. Find the common ratio and write out the first four terms of the geometric sequence
{8^n-2/10}
its common difference is
its first term is
its second term is
its third term is
its fourth term is

3. For the arithmetic sequence with given first term 3 and common difference 4
its nth term is
its 10th term is

Answer 1

1. Each term is the sum of the previous two terms. So if the first two terms are 2, 3...
the third is 5 (sum of 2+3), the fourth is 8 (sum of 3+5), the fifth is 13 (the sum of 5+8)

2. The common ratio of a geometric sequence is the ratio of any two adjacent terms. When n increases by one, in the equation (8^n-2/10), 8 is raised to one higher power, so each term is 8 times the previous one. Assuming that the sequence is 8^(n-2)/10, the first four terms are: 1/80, 1/10, 8/10, 64/10

3. The common difference of an arithmetic sequence is the difference between from each term to the next one. So if the first term is 3, and the common difference is 4, the second is 7, the third is 11, etc. The Nth term is 4(n-1)+3 (where increasing n by 1 adds four, and when n=1 its value is 3), or 4n-1, and the 10th term is 4*10-1 = 39.

Answer 2

Don't panic!

OK, for the first one the nth term is equal to the n-1st term plus the n-2nd term, or the sum of the previous two terms. This actually is the famous Fibonacci series...anyway, the third term is the sum of the first and second; the fourth is the sum of the third and second; the fifth is the sum of the fourth and third. The actual Fibonacci series is thus 1,1,2,3,5,8

For the second one: the first term is 8^(0-2)/10 and the second is 8^(1-2)/10, or 1/640 and 1/80. The third one is 8^(2-2)/10, or 1/10, and the fourth is 8^(3-2)/10, or 80. And so on--this is assuming that you're starting with n=0. The common ratio of this sequence is 8.

For the third one, a common difference of 4 means that you just add four each time to get the next number, so the sequence is 3,7,11,15... the 10th term is 39 if I've done the math right.