Problem: Given a geometric progression of five terms, each a positive integer less than 100. The sum of the five terms is 211. If S is the sum of those terms in the progression which are squares of integers, then S is:
Answer Choices:
A. 0
B. 91
C. 133
D. 195
E. 211
Solution:
S=a(1+r+r2+r3+r4)= 211. If r is an integer a must equal 1 since 211 is prime. However, neither a=1,r=1 nor a=1,r=2, nor a=1,r=3 is satisfactory. ∴2<r<3 or 1<r<2. Noting that when r=nm,m,n integers, a must equal n4 since 211 is prime, we tryr=23,a=16. ∴s=16(1+23+(23)2+(23)3+(23)4)=211, and the combination a=16,r=23 is usable. By symmetry the combination a=81,r=32 is also usable. In either case, the odd-numbered terms are squares of integers, and their sum is 16+36+81=133.