Geometric Sequences

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Geometric Sequences · Grade 12 · Worksheet 2

1. A geometric sequence has its first three terms represented by the areas of three concentric circles. The smallest circle has radius 2 cm, and each subsequent circle's radius increases by a constant factor. If the sum of the areas of the first three circles is 84π cm², what is the common ratio of the geometric sequence formed by the areas? Answer: 4 Solution: We have three concentric circles with radii in geometric progression (constant ratio for the radii). The smallest radius is 2 cm. Let the common ratio of the radii be \( r \).
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understand the problem** We have three concentric circles with radii in geometric progression (constant ratio for the radii). The smallest radius is 2 cm. Let the common ratio of the radii be \( r \). Then the radii are: \( R_1 = 2 \) \( R_2 = 2r \) \( R_3 = 2r^2 \) The areas of the circles are: \( A_1 = \pi (2)^2 = 4\pi \) \( A_2 = \pi (2r)^2 = \pi (4r^2) = 4\pi r^2 \) \( A_3 = \pi (2r^2)^2 = \pi (4r^4) = 4\pi r^4 \) These areas \( A_1, A_2, A_3 \) form a geometric sequence themselves. --- **Step 2: Condition given** Sum of the areas of the first three circles = \( 84\pi \) cm². So: \( 4\pi + 4\pi r^2 + 4\pi r^4 = 84\pi \) --- **Step 3: Simplify the equation** Divide through by \( 4\pi \): \( 1 + r^2 + r^4 = 21 \) So: \( r^4 + r^2 + 1 = 21 \) \( r^4 + r^2 - 20 = 0 \) --- **Step 4: Substitute \( x = r^2 \)** Let \( x = r^2 \). Then \( r^4 = x^2 \). Equation becomes: \( x^2 + x - 20 = 0 \) Factor: \( (x + 5)(x - 4) = 0 \) So \( x = -5 \) or \( x = 4 \). Since \( x = r^2 \) and \( r > 0 \), we take \( x = 4 \). Thus \( r^2 = 4 \), so \( r = 2 \) (common ratio of the radii). --- **Step 5: Find the common ratio of the geometric sequence of areas** The areas are: \( A_1 = 4\pi \) \( A_2 = 4\pi r^2 = 4\pi \times 4 = 16\pi \) \( A_3 = 4\pi r^4 = 4\pi \times 16 = 64\pi \) Check: \( 4\pi + 16\pi + 64\pi = 84\pi \) ✔ The areas sequence: \( 4\pi, 16\pi, 64\pi \) Common ratio of areas = \( \frac{16\pi}{4\pi} = 4 \), or \( \frac{64\pi}{16\pi} = 4 \). --- **Final Answer:** 4

2. ∑(n=1 to 5) 5 × (-4)^(n-1) = ? Answer: 1025 Solution: a₁ = 5 × (-4)^(1-1) = 5 × (-4)^0 = 5 × 1 = 5 r = -4 Sₙ = a₁ × (1 - rⁿ) / (1 - r) Substitute n = 5, a₁ = 5, r = -4 S₅ = 5 × (1 - (-4)⁵) / (1 - (-4)) Calculate (-4)⁵ (-4)⁵ = -1024 S₅ = 5 × (1 - (-1024)) / (1 + 4) S₅ = 5 × (1 + 1024) / 5 S₅ = 5 × 1025 / 5 S₅ = 1025 The answer is 1025.
   Full step-by-step solution

   Step 1: Identify the first term (a₁) and common ratio (r) a₁ = 5 × (-4)^(1-1) = 5 × (-4)^0 = 5 × 1 = 5 r = -4 Step 2: Use the geometric series sum formula for n terms Sₙ = a₁ × (1 - rⁿ) / (1 - r) Step 3: Substitute n = 5, a₁ = 5, r = -4 S₅ = 5 × (1 - (-4)⁵) / (1 - (-4)) Step 4: Calculate (-4)⁵ (-4)⁵ = -1024 Step 5: Substitute into the formula S₅ = 5 × (1 - (-1024)) / (1 + 4) S₅ = 5 × (1 + 1024) / 5 S₅ = 5 × 1025 / 5 S₅ = 1025 The answer is 1025.

3. Tane is a forestry engineer monitoring the growth of a rare tree species in a conservation plot. He records that the height of a particular tree follows a geometric progression. The tree was 7 cm tall when first measured. Each subsequent year, its height increases by a factor of 3 times the previous year's growth increment (not the total height). In the first year, it grew 7 cm; in the second year, it grew 21 cm; in the third year, it grew 63 cm, and so on. Tane needs to report the total height of the tree after 9 years (the sum of all growth increments from year 1 to year 9) and the height it reached at the end of the 9th year (the 9th term of the sequence of growth increments). Find the growth increment in the 9th year and the total height of the tree after 9 years. Answer: a9 = 45927, S9 = 68887 Solution: Identify the geometric sequence of growth increments: a1 = 7, common ratio r = 3, number of terms n = 9.
   Full step-by-step solution

   Step 1: Identify the geometric sequence of growth increments: a1 = 7, common ratio r = 3, number of terms n = 9. Step 2: Find the 9th term (growth increment in year 9) using a_n = a1 * r^(n-1): a9 = 7 * 3^(9-1) = 7 * 3^8 3^8 = 3^4 * 3^4 = 81 * 81 = 6561 a9 = 7 * 6561 = 45927 Step 3: Find the total height after 9 years using S_n = a1 * (r^n - 1) / (r - 1): S9 = 7 * (3^9 - 1) / (3 - 1) 3^9 = 3 * 3^8 = 3 * 6561 = 19683 S9 = 7 * (19683 - 1) / 2 S9 = 7 * 19682 / 2 S9 = 7 * 9841 = 68887 Step 4: State the results: The growth increment in the 9th year is 45927 cm, and the total height after 9 years is 68887 cm.

4. Emma is a biologist tracking the growth of a bacterial colony in a laboratory experiment. She observes that the colony's population follows a geometric pattern. At the start of the experiment (day 1), the colony has 50 bacteria. Each subsequent day, the population increases by a factor of 4 times the previous day's population. Emma needs to predict the population after 5 days (the 5th term) and the total number of bacteria that have existed over the first 5 days (including day 1). Find the population on day 5 and the total population over the first 5 days. Answer: a5 = 12800, S5 = 17050 Solution: Identify the first term a1 = 50 and the common ratio r = 4. The number of terms n = 5. Use the nth term formula a_n = a1 * r^(n-1).
   Full step-by-step solution

   Step 1: Identify the first term a1 = 50 and the common ratio r = 4. The number of terms n = 5. Step 2: Use the nth term formula a_n = a1 * r^(n-1). For n=5, a5 = 50 * 4^(5-1) = 50 * 4^4. Step 3: Calculate 4^4 = 4 * 4 * 4 * 4 = 256. Step 4: Multiply by 50: a5 = 50 * 256 = 12800. The population on day 5 is 12800 bacteria. Step 5: Use the sum formula S_n = a1 * (r^n - 1) / (r - 1). For n=5, S5 = 50 * (4^5 - 1) / (4 - 1). Step 6: Calculate 4^5 = 4 * 4^4 = 4 * 256 = 1024. Step 7: Numerator: 1024 - 1 = 1023. Denominator: 4 - 1 = 3. Step 8: Divide: 1023 / 3 = 341. Step 9: Multiply by 50: S5 = 50 * 341 = 17050. The total population over the first 5 days is 17050 bacteria. Step 10: Final answer: a5 = 12800, S5 = 17050.