(h) The sum of the first four terms of a geometric series is equal to 45 and the sum of the first eight terms is 765. Determine the first 3 terms of the series.

Let’s determine the first three terms of the geometric series given the following information: 1. **Sum of the first four terms (S₄):** \[ S_4 = a + ar + ar^2 + ar^3 = 45 \] Using the formula for the sum of the first \( n \) terms of a geometric series: \[ S_n = a \frac{1 - r^n}{1 - r} \] So, \[ 45 = a \frac{1 - r^4}{1 - r} \quad \text{(1)} \] 2. **Sum of the first eight terms (S₈):** \[ S_8 = a + ar + ar^2 + ar^3 + ar^4 + ar^5 + ar^6 + ar^7 = 765 \] Using the sum formula: \[ 765 = a \frac{1 - r^8}{1 - r} \quad \text{(2)} \] **Dividing Equation (2) by Equation (1):** \[ \frac{765}{45} = \frac{a \frac{1 - r^8}{1 - r}}{a \frac{1 - r^4}{1 - r}} \implies 17 = \frac{1 - r^8}{1 - r^4} \] Notice that \( 1 - r^8 = (1 - r^4)(1 + r^4) \), so: \[ 17 = \frac{(1 - r^4)(1 + r^4)}{1 - r^4} \implies 1 + r^4 = 17 \implies r^4 = 16 \implies r = \pm 2 \] **Case 1: \( r = 2 \)** \[ 45 = a \frac{1 - 16}{1 - 2} = a \frac{-15}{-1} = 15a \implies a = 3 \] First three terms: \[ 3, \quad 3 \times 2 = 6, \quad 6 \times 2 = 12 \] **Case 2: \( r = -2 \)** \[ 45 = a \frac{1 - 16}{1 + 2} = a \frac{-15}{3} = -5a \implies a = -9 \] First three terms: \[ -9, \quad -9 \times (-2) = 18, \quad 18 \times (-2) = -36 \] **Conclusion:** The first three terms of the series can either be \( 3, 6, 12 \) or \( -9, 18, -36 \). Assuming a positive common ratio, the first three terms are: **Answer:** The first three terms are 3, 6, and 12.