Mathematics Ma Lecture Notes - Lecture 4: Geometric Progression

• School: Harvard University
• Department: Mathematics
• Course: Mathematics Ma

Document Summary

Problem involving geometric sequence: the first four terms of a geometric progression are 1,x,2x,y. Firstly, a geometric sequence has common ratio ; that is; they have a common multiplier that can divide through each terms without a reminder just like multiplying through by the ratio 1/r. Hence; to find x and y; we need to find the common ratio. If the first term let"s a=1, 2nd term=x, 3rd term=2x and the fourth term=y. Provided that they have a common ratio rn-1 where n=number of term. Then let"s solve for the first term; a=r1-1=r0 r0=1 (remember indices) a=r0=1. Let"s find x now; by substituting x=r in equation 1 into equation ii above to find r; Therefore, y=8 and x=2: the third and fifth terms of a geometric progression are -8 and -72. Find the possible values of the twentieth term of the progression. Recall what we said about geometric sequence earlier, presence of common ratio;