Curious, different mistakes in #1 and #2 above, both leading to the same incorrect answer.
Suppose that the first series is
\(\displaystyle a, \ ar,\ ar^{2}, \ ar^{3}, \dots\)
Its sum to infinity is 2000, so
\(\displaystyle \frac{a}{1-r}=2000\dots(1)\)
For the second series, each term of the first series is squared, so the second series will be
\(\displaystyle a^{2}, \ a^{2}r^{2}, \ a^{2}r^{4}, \ a^{2}r^{6}, \dots\)
Its sum to infinity is 16 times the sum of the first series, so
\(\displaystyle \frac{a^{2}}{1-r^{2}}=32000\dots(2)\)
Squaring equation (1) and using that to eliminate the a^2 term leads, after some simplification, to the quadratic
\(\displaystyle 63r^{2} - 125r +62 = 0,\)
from which
\(\displaystyle (63r-62)(r-1)=0,\)
so r = 62/63, (r can't equal 1).
Substitution in (1) shows that a = 2000/63.
