Problem:
Two distinct, real, infinite geometric series each have a sum of 1 and have the same second term. The third term of one of the series is 1/8, and the second term of both series can be written in the form pm​−n​, where m,n, and p are positive integers and m is not divisible by the square of any prime. Find 100m+10n+p.
Solution:
Let the two series be
k=0∑∞​a⋅rk and k=0∑∞​b⋅sk
The given conditions imply that a=1−r,b=1−s, and ar=bs. It follows that r(1−r)=s(1−s), that is r−s=r2−s2. Because the series are not identical, rî€ =s, leaving r=1−s as the only possibility, and the series may be written as
k=0∑∞​(1−r)⋅rk and k=0∑∞​r⋅(1−r)k
As we may pick either series as the one whose third term is 1/8, set (1−r)r2=1/8, from which we obtain 8r3−8r2+1=0. The substitution t=2r yields t3−2t2+1=0, for which 1 is a root. Factoring gives (t−1)(t2−t−1)=0, so the other two roots are (1±5​)/2, which implies that r=1/2 or r=(1±5​)/4. However, if r were 1/2, the two series would be equal; and if r were (1−5​)/4, then s would be (3+5​)/4, but convergence requires that ∣s∣<1. Thus r=(1+5​)/4, and −1<s=(3−5​)/4<1. The second term of the series is therefore equal to