Problem:
The geometric series a+ar+ar2+⋯ has a sum of 7 , and the terms involving odd powers of r have a sum of 3 . What is a+r ?
Answer Choices:
A. 34
B. 712
C. 23
D. 37
E. 25
Solution:
The terms involving odd powers of r form the geometric series ar+ar3+ar5+⋯. Thus
7=a+ar+ar2+⋯=1−ra
and
3=ar+ar3+ar5+⋯=1−r2ar=1−ra⋅1+rr=1+r7r
Therefore r=3/4. It follows that a/(1/4)=7, so a=7/4 and
a+r=47+43=25
OR
The sum of the terms involving even powers of r is 7−3=4. Therefore
3=ar+ar3+ar5+⋯=r(a+ar2+ar4+⋯)=4r
so r=3/4. As in the first solution, a=7/4 and a+r=5/2.
The problems on this page are the property of the MAA's American Mathematics Competitions