Problem:
The sequences of positive integers 1,a2,a3,… and 1,b2,b3,… are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let cn=an+bn. There is an integer k such that ck−1=100 and ck+1=1000. Find ck.
Solution:
Because c1=2<100, it follows that k−1≥2, so k≥3. There are integers d≥0 and r≥1 such that an=1+(n−1)d and bn=rn−1, so 100=ck−1= 1+(k−2)d+rk−2 and 1000=ck+1=1+kd+rk. Subtracting these equations gives 900=rk−rk−2+2d=rk−3(r−1)r(r+1)+2d. Because (r−1)r(r+1) must be a multiple of 3,d is also a multiple of 3. Because 100=1+(k−2)d+rk−2, it follows that r is also a multiple of 3. The restrictions rk−2≤99 and rk≤999 show that (r,k) must be one of (3,3),(3,4),(3,5),(3,6),(6,3), or (9,3). For the first five of these there is no integer value for d that satisfies all the required conditions, but if r=9 and k=3, then d=90 does satisfy all the required conditions. In this case ck=1+(3−1)90+93−1=262.
The problems on this page are the property of the MAA's American Mathematics Competitions