(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.
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To solve this problem, let’s denote the first term of the geometric series as \( a \) and the common ratio as \( r \). The sum of the first four terms is given by the formula: \[ S_4 = a + ar + ar^2 + ar^3 = a(1 + r + r^2 + r^3) \] The sum of the first eight terms is: \[ S_8 = a + ar + ar^2 + ar^3 + ar^4 + ar^5 + ar^6 + ar^7 = a(1 + r + r^2 + r^3 + r^4 + r^5 + r^6 + r^7) \] We have two equations: 1. \( S_4 = a(1 + r + r^2 + r^3) = 45 \) 2. \( S_8 = a(1 + r + r^2 + r^3 + r^4 + r^5 + r^6 + r^7) = 765 \) Notice that \( S_8 \) can also be expressed in terms of \( S_4 \): \[ S_8 = S_4 + ar^4(1 + r + r^2 + r^3) = S_4 + ar^4 S_4 = S_4(1 + r^4) \] Substituting for \( S_4 \): \[ 765 = 45(1 + r^4) \] Dividing both sides by 45: \[ 17 = 1 + r^4 \] This simplifies to: \[ r^4 = 16 \implies r = 2 \] (We take the positive root since negative ratios would not keep the signs consistent in the series.) Now substituting \( r = 2 \) back into the first equation to find \( a \): \[ S_4 = a(1 + 2 + 4 + 8) = 15a = 45 \implies a = 3 \] The first three terms of the series are: 1. \( a = 3 \) 2. \( ar = 3 \cdot 2 = 6 \) 3. \( ar^2 = 3 \cdot 2^2 = 12 \) So, the first three terms of the geometric series are: **3, 6, 12.**
