Problem:
A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a fourterm geometric sequence of positive integers. The first three terms of the resulting four-term sequence are 57,60, and 91. What is the fourth term of this sequence?
Answer Choices:
A. 190
B. 194
C. 198
D. 202
E. 206
Solution:
Let the terms of the arithmetic sequence be a0,a1,a2,a3, and let the terms of the geometric sequence be g0,g1,g2,g3. There must be constants b,d,c, and r such that an=b+d⋅n and gn=c⋅rn for n=0,1,2,3. Let the terms of the sum sequence be xn=an+gn. It is given that x0=57,x1=60, and x2=91. With the goal of eliminating two of the constants, consider x2−2x1+x0. Then
28=91−2⋅60+57=(b+2d+cr2)−2(b+d+cr)+(b+c)=c(r−1)2
Because g1 and g2 are integers, r must be a rational number, say qp for some relatively prime positive integers p and q. Then the equation above becomes c(p−q)2=28q2. Because gcd(p−q,q)= gcd(p,q)=1, either (p−q)2=4 or (p−q)2=1.
If (p−q)2=1, then c=28q2. Because c<57, it must be that q=1 and c=28. Then p−q=1 (because p=0 ), so p=2 and r=2. It follows that b=x0−c=57−28=29, d=a1−b=(60−2⋅28)−29=−25, and a2=29+2⋅(−25)<0, violating the conditions of the problem.
Therefore (p−q)2=4 and c=7q2. Because p−q is even, q must be odd and the only choice that makes c<57 is q=1. Then c=7,p=3, and r=3. It follows that the geometric sequence is 7,21,63,189. Therefore a0=57−7=50 and a1=60−21=39, giving d=−11. Therefore a3=50+3(−11)=17 and x3=17+189=(E)206.
The problems on this page are the property of the MAA's American Mathematics Competitions