Note: You must be registered in order to post a reply. To register, click here. Registration is FREE!
T O P I C R E V I E W
Posted - 01/03/2011 : 19:25:50 Suppose that a, b, and c are three consecutive terms in a geometric sequence. Show that 1/(a+b) , 1/(2b) , and 1/(c+b) are three consecutive terms in an arithmetic sequence.
I have no clue how to approach this question! It's from my textbook, so i know all relevant equations/formulas. Please help me out, my teacher is very strict.
EDIT: I have a second question:
Let b denote a positive constant. Find the sum of the first n terms in the sequence
1/(1+b) , 1/(1-b) , 1/(1-b) , ...
I get that the denominators must form some kind of arithmetic sequence, but I can't for the life of me figure out what the common difference could be... please help!!
1 L A T E S T R E P L I E S (Newest First)
Posted - 01/04/2011 : 19:22:40 Since the sequence is geometric there is a common ratio r. Then b=ar and c= ar. See if you can take it from there.