Without any more information, the first term can be any number.
Let a be the first term in the geometric progression, and let r be the common ratio between consecutive terms. Then
a + ar + ar ² + ar ³ + ar ⁴ + ar ⁵ = 9 (a + ar + ar ²)
ar ³ + ar ⁴ + ar ⁵ = 8 (a + ar + ar ²)
r ³ + r ⁴ + r ⁵ = 8 (1 + r + r ²)
r ³ (1 + r + r ²) = 8 (1 + r + r ²)
r ³ = 8
r = 2
Now with this ratio, the sum of the first six terms is
a (1 + r + r ² + r ³ + r ⁴ + r ⁵) = a (1 - r ⁶)/(1 - r) = 63a
while the sum of the first three terms is
a (1 + r + r ²) = a (1 - r ³)/(1 - r) = 7a
and of course 63a = 9 • 7a, and this is true for any a.
