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Consider the following geometric sequence: 1 024 ; 256 ; 64 ; .. - NSC Mathematics - Question 3 - 2022 - Paper 1
Question 3
Consider the following geometric sequence: 1 024 ; 256 ; 64 ; ... Calculate: 3.1.1 The 10th term of the sequence 3.1.2 \( \sum_{p=0}^{9} 256(4^{-p}) \) The first... show full transcript
Worked Solution & Example Answer:Consider the following geometric sequence: 1 024 ; 256 ; 64 ; .. - NSC Mathematics - Question 3 - 2022 - Paper 1
Step 1
3.1.1 The 10th term of the sequence
Answer
To find the 10th term of a geometric sequence, we can use the formula:
[ T_n = ar^{n-1} ]
where:
- ( a = 1024 ) (the first term)
- ( r = \frac{256}{1024} = \frac{1}{4} ) (common ratio)
- ( n = 10 )
Substituting the values:
[ T_{10} = 1024 \left(\frac{1}{4}\right)^{10-1} = 1024 \left(\frac{1}{4}\right)^{9} = \frac{1024}{262144} = \frac{1}{256} ]
Step 2
3.1.2 \( \sum_{p=0}^{9} 256(4^{-p}) \)
Answer
The formula for the sum of the first ( n ) terms of a geometric series is:
[ S_n = \frac{a(1 - r^n)}{1 - r} ]
Here,
- ( a = 256 )
- ( r = \frac{1}{4} )
- ( n = 10 )
Calculating:
[ S_{10} = \frac{256(1 - (\frac{1}{4})^{10})}{1 - \frac{1}{4}} ]
[ = \frac{256(1 - \frac{1}{1048576})}{\frac{3}{4}} ]
Simplifying:
[ S_{10} = \frac{256 \cdot \frac{3}{4}(1048575)}{1048576} \approx 365.33 ]
Step 3
3.2 Determine the values of \( r \) for which the sequence will converge.
Answer
To determine the convergence of a geometric sequence, the absolute value of the common ratio ( r ) must be less than 1:
[ |r| < 1 ]
From the provided terms:
- First term: ( -r^{2} - 6r - 9 )
- Second term: ( \frac{r^{3} + 9r^{2} + 27r + 27}{2} )
Equating the terms:
[ r^{2} + 6r + 9 = 0 ]
Factoring gives us:
[ (r+3)^{2} = 0 ]
So, ( r = -3 ). Check convergence:
[ -1 < r < 1 \Rightarrow -1 < -3 < 1 ] (this does not satisfy the convergence, thus it diverges). Hence, the sequence converges for: [ -1 < r < 1 ]