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 a0β,a1β,a2β,a3β, and let the terms of the geometric sequence be g0β,g1β,g2β,g3β. There must be constants b,d,c, and r such that anβ=b+dβ
n and gnβ=cβ
rn for n=0,1,2,3. Let the terms of the sum sequence be xnβ=anβ+gnβ. It is given that x0β=57,x1β=60, and x2β=91. With the goal of eliminating two of the constants, consider x2ββ2x1β+x0β. Then
28=91β2β
60+57=(b+2d+cr2)β2(b+d+cr)+(b+c)=c(rβ1)2
Because g1β and g2β are integers, r must be a rational number, say qpβ for some relatively prime positive integers p and q. Then the equation above becomes c(pβq)2=28q2. Because gcd(pβq,q)= gcd(p,q)=1, either (pβq)2=4 or (pβq)2=1.
If (pβq)2=1, then c=28q2. Because c<57, it must be that q=1 and c=28. Then pβq=1 (because pξ =0 ), so p=2 and r=2. It follows that b=x0ββc=57β28=29, d=a1ββb=(60β2β
28)β29=β25, and a2β=29+2β
(β25)<0, violating the conditions of the problem.
Therefore (pβq)2=4 and c=7q2. Because pβq 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 a0β=57β7=50 and a1β=60β21=39, giving d=β11. Therefore a3β=50+3(β11)=17 and x3β=17+189=(E)206β.
The problems on this page are the property of the MAA's American Mathematics Competitions