\( a_{n}=8192 / 2187 \) in the geometric sequence with \( a_{1}=2 / 3 \) and \( a_{4}=-128 / 81 \) \( a_{41}=a_{1}(r)^{3} \)

To unravel this geometric puzzle, let's first determine the common ratio \( r \). We can use the relationship between the first term and the fourth term: \[ a_4 = a_1 r^3. \] We know \( a_1 = \frac{2}{3} \) and \( a_4 = -\frac{128}{81} \), which gives us: \[ -\frac{128}{81} = \frac{2}{3} r^3. \] Multiplying both sides by \( \frac{3}{2} \) to solve for \( r^3 \): \[ r^3 = -\frac{128}{81} \cdot \frac{3}{2} = -\frac{128 \cdot 3}{81 \cdot 2} = -\frac{384}{162} = -\frac{128}{54}. \] Calculating \( r \): \[ r = \sqrt[3]{-\frac{128}{54}} = \sqrt[3]{-\frac{64}{27}} = -\frac{4}{3}. \] Now, we can find \( a_{41} \): \[ a_{41} = a_1 r^{40} = \frac{2}{3} \left(-\frac{4}{3}\right)^{40}. \] Since the exponent is even, this evaluates to: \[ a_{41} = \frac{2}{3} \cdot \frac{4^{40}}{3^{40}} = \frac{2 \cdot 4^{40}}{3^{41}}. \] And with \( 4^{40} = (2^2)^{40} = 2^{80} \): \[ a_{41} = \frac{2^{81}}{3^{41}}. \] This result captures the exciting growth of the geometric sequence!