In an infinite GP converging to a finite limit, S=a/(1-r) because Sn=a(r^n-1)/(r-1)=a(1-r^n)/(1-r). When r is small (<1) and n is large r^n approaches zero and is zero when n=infinity. Therefore, 5/2=a/(1-r) so 5(1-r)=2a. This shows a relationship between a and r for 0<r<1. This equation is 5-5r=2a or a=5(1-r)/2. If this is graphed for the domain 0<r<1, it shows the values a can take. Examples: r=1/2, a=5/4; r=1/3, a=5/3; r=1/5, a=2. So there are infinite solutions for (r,a). r can be negative: -1<r<1 is the whole domain for r. Example: r=-1/5, a=3.