2.3. The first three terms of a geometric sequence has a sum of 21 and their product is 64 . Determine the value of the first term, if the common ratio is an integer, i.e. \( r \in \mathbb{Z} \).

In a geometric sequence, the first three terms can be represented as \( a \), \( ar \), and \( ar^2 \). Given that their sum is 21, we have: \[ a + ar + ar^2 = 21 \] \[ a(1 + r + r^2) = 21 \tag{1} \] The product of the first three terms is given as 64: \[ a \cdot ar \cdot ar^2 = 64 \] \[ a^3 r^3 = 64 \] \[ a^3 r^3 = 4^3 \] \[ (ar)^3 = 64 \tag{2} \] From equation (2), we can isolate \( ar \): \[ ar = 4 \] \[ a = \frac{4}{r} \] Now, substitute \( a = \frac{4}{r} \) into equation (1): \[ \frac{4}{r}(1 + r + r^2) = 21 \] \[ 4(1 + r + r^2) = 21r \] \[ 4 + 4r + 4r^2 = 21r \] \[ 4r^2 - 17r + 4 = 0 \] Now we can solve this quadratic equation using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{17 \pm \sqrt{(-17)^2 - 4 \cdot 4 \cdot 4}}{2 \cdot 4} \] \[ = \frac{17 \pm \sqrt{289 - 64}}{8} = \frac{17 \pm \sqrt{225}}{8} \] \[ = \frac{17 \pm 15}{8} \] This gives us: \[ r = \frac{32}{8} = 4 \quad \text{or} \quad r = \frac{2}{8} = \frac{1}{4} \] However, since \( r \) must be an integer, we have \( r = 4 \). Now we can find \( a \): \[ a = \frac{4}{r} = \frac{4}{4} = 1 \] Thus, the value of the first term \( a \) is \( 1 \). **Final Answer:** The value of the first term \( a \) is \( 1 \).