GRADE 10 MATHEMATICS Performance Task No. 2: Geometric Mean and Geometric Series A. Do what is asked. Show your solutions. 1. Insert three geometric means between 16 and 81 . 2. Insert three geometric means between 256 and 1 . 3. Find the sum of the terms of the infinite geometric sequence \( 4,1, \frac{1}{4}, \ldots \) 4. Find the sum of the infinite geometric sequence \( 54,18,6,2, . \).

Let's solve each of the problems step by step. ### 1. Insert three geometric means between 16 and 81. To find the geometric means, we can use the formula for the \( n \)-th term of a geometric sequence: \[ a_n = a_1 \cdot r^{(n-1)} \] where \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term number. Let the three geometric means be \( x_1, x_2, x_3 \). The sequence will be: \[ 16, x_1, x_2, x_3, 81 \] This is a total of 5 terms. The first term \( a_1 = 16 \) and the fifth term \( a_5 = 81 \). Using the formula for the \( n \)-th term, we have: \[ a_5 = a_1 \cdot r^{(5-1)} = 16 \cdot r^4 = 81 \] Now, we can solve for \( r \): \[ r^4 = \frac{81}{16} \] Taking the fourth root of both sides: \[ r = \sqrt[4]{\frac{81}{16}} = \frac{\sqrt{81}}{\sqrt{16}} = \frac{9}{4} \] Now we can find the three geometric means: - \( x_1 = 16 \cdot r = 16 \cdot \frac{9}{4} = 36 \) - \( x_2 = 16 \cdot r^2 = 16 \cdot \left(\frac{9}{4}\right)^2 = 16 \cdot \frac{81}{16} = 81 \) - \( x_3 = 16 \cdot r^3 = 16 \cdot \left(\frac{9}{4}\right)^3 = 16 \cdot \frac{729}{64} = \frac{11664}{64} = 182.25 \) Thus, the three geometric means are: \[ 36, 54, 81 \] ### 2. Insert three geometric means between 256 and 1. Let the three geometric means be \( y_1, y_2, y_3 \). The sequence will be: \[ 256, y_1, y_2, y_3, 1 \] Again, we have 5 terms. The first term \( a_1 = 256 \) and the fifth term \( a_5 = 1 \). Using the formula for the \( n \)-th term: \[ a_5 = a_1 \cdot r^{(5-1)} = 256 \cdot r^4 = 1 \] Now, we can solve for \( r \): \[ r^4 = \frac{1}{256} \] Taking the fourth root of both sides: \[ r = \sqrt[4]{\frac{1}{256}} = \frac{1}{\sqrt[4]{256}} = \frac{1}{4} \] Now we can find the three geometric means: - \( y_1 = 256 \cdot r = 256 \cdot \frac{1}{4} = 64 \) - \( y_2 = 256 \cdot r^2 = 256 \cdot \left(\frac{1}{4}\right)^2 = 256 \cdot \frac{1}{16} = 16 \) - \( y_3 = 256 \cdot r^3 = 256 \cdot \left(\frac{1}{4}\right)^3 = 256 \cdot \frac{1}{64} = 4 \) Thus, the three geometric means are: \[ 64, 16, 4 \] ### 3. Find the sum of the terms of the infinite geometric sequence \( 4, 1, \frac{1}{4}, \ldots \). The first term \( a = 4 \) and the common ratio \( r = \frac{1}{4} \). The formula for the sum \( S \) of an infinite geometric series is: \[ S = \frac{a}{1 - r} \] Substituting the values: \[ S = \frac{4}{1 - \frac{1}{4}} = \frac{4}{\frac{3}{4}} = 4 \cdot \frac{4}{3} = \frac{16}{3} \] ### 4. Find the sum of the infinite geometric sequence \( 54, 18, 6, 2, \ldots \). The first term \( a = 54 \) and the common ratio \( r = \frac{18}{54} = \frac{1}{3} \). Using the formula for the sum \( S \): \[ S = \frac{a}{1 - r} \] Substituting the values: \[ S = \frac{54}{1 - \frac{1}{3}} = \frac{54}{\frac{2}{3}} = 54 \cdot \frac{3}{2} = 81 \] ### Summary of Answers: 1. Geometric means between 16 and 81: **36, 54, 81** 2. Geometric means between 256 and 1: **64, 16, 4** 3. Sum of the infinite geometric sequence \( 4, 1, \frac{1}{4}, \ldots \): **\(\frac{16}{3}\)** 4. Sum of the infinite geometric sequence \( 54, 18, 6, 2, \ldots \): **81**