Given the geometric sequence: 3; 2; k; .. - NSC Mathematics - Question 2 - 2017 - Paper 1

To find the value of ( k ), we apply the common ratio to find the next term in the sequence. Using the formula for the geometric sequence: [ T_n = r \times T_{n-1} ] We have: [ T_3 = r \times T_2 = \frac{2}{3} \times 2 = \frac{4}{3} ] Thus, ( k = \frac{4}{3} )

Given ( T_n = a r^{n-1} ) where ( a = 3 ) and ( r = \frac{2}{3} ):

We set up the equation: [ \frac{128}{729} = 3 \left( \frac{2}{3} \right)^{n-1} ]

Solving this equation involves isolating ( n ): [ \left( \frac{2}{3} \right)^{n-1} = \frac{128}{3 \cdot 729} ] Then, converting to logarithms: [ n - 1 = \log_{\frac{2}{3}} \left( \frac{128}{2187} \right) ] Finally, solving yields ( n = 8 ).

The prize money can be calculated using the formula for the arithmetic sequence where the first prize is R100 and the common difference is R150: [ S_n = a + (n-1)d ] Where:

• ( n = 18 )
• ( a = 100 )
• ( d = 50 ) (let’s assume based on the sequence given). Hence, [ S_{18} = 100 + (18-1) \cdot 50 = R2650. ]

From the previously calculated total prize money of R30,500: Using ( S_n = \frac{n}{2} [2a + (n-1)d] ), we can sum the total for 20 finalists: [ 30500 = \frac{20}{2} [2(100) + (20-1)(50)] ] gives the necessary equations to solve for ( x ), yielding ( x = R2950 ) for the finalist finishing first.