Let a(n) denote the n-th term in the GP. We're given
a(6) = 8a(10)
a(7) + a(8) = 192
In a geometric progression, consecutive terms are scaled up or down by a fixed ratio r such that
a(n) = r a(n - 1)
By substitution, we get
a(n) = r (r a(n - 2)) = r ² a(n - 2)
and we can continue the pattern down to the first term,
a(n) = r a(n - 1) = r ² a(n - 2) = r ³ a(n - 3) = … = rⁿ ⁻¹ a(1)
So we can rewrite the first two equations in terms of the first term of the GP a(1) and the common ratio r :
r ⁵ a(1) = 8r ⁹ a(1)
r ⁶ a(1) + r ⁷ a(1) = 192
Solve the first equation for r :
r ⁵ a(1) = 8r ⁹ a(1) → 1 = 8r ⁴ → r = 1/∜8 ≈ 0.5946
Solve the second equation for a(1) :
r ⁶ a(1) + r ⁷ a(1) = r ⁶ a(1) (1 + r ) = 1/(16√2) a(1) (1 + 1/∜8) = 192
→ a(1) = 6144 ∜2 / (1 + ∜8) ≈ 2724.48