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Given the quadratic sequence: 0; 17; 32; .. - NSC Mathematics - Question 3 - 2017 - Paper 1 Question 3
View full question Given the quadratic sequence: 0; 17; 32; ...
3.1 Determine an expression for the general term, $T_n$, of the quadratic sequence.
3.2 Which terms in the quadratic s... show full transcript
View marking scheme Worked Solution & Example Answer:Given the quadratic sequence: 0; 17; 32; .. - NSC Mathematics - Question 3 - 2017 - Paper 1
Determine an expression for the general term, $T_n$, of the quadratic sequence. Only available for registered users.
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To find the general term of the quadratic sequence, we first compute the first and second differences:
From the given terms: 0, 17, 32,
First differences: 17 - 0 = 17, 32 - 17 = 15 → Results: 17, 15
Second differences: 15 - 17 = -2 → Constant second difference of -2.
Using the polynomial form for a quadratic sequence:
T n = a n 2 + b n + c T_n = an^2 + bn + c T n = a n 2 + bn + c
The second difference is given by 2 a = − 2 2a = -2 2 a = − 2 , thus a = − 1 a = -1 a = − 1 .
The first differences can be expressed as:
3 a + b = 17 3a + b = 17 3 a + b = 17
Substituting a = − 1 a = -1 a = − 1 gives:
ightarrow b = 20$$
Now using the initial term, where n = 1 n=1 n = 1 ,
ightarrow c = -19$$
Hence, the general term is:
T n = − n 2 + 20 n − 19 T_n = -n^2 + 20n - 19 T n = − n 2 + 20 n − 19
Which terms in the quadratic sequence have a value of 56? Only available for registered users.
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To find the terms in the sequence that equal 56:
Set the general term equal to 56:
56 = − n 2 + 20 n − 19 56 = -n^2 + 20n - 19 56 = − n 2 + 20 n − 19
Rearranging gives:
n 2 − 20 n + 75 = 0 n^2 - 20n + 75 = 0 n 2 − 20 n + 75 = 0
Factoring the equation:
( n − 15 ) ( n − 5 ) = 0 (n - 15)(n - 5) = 0 ( n − 15 ) ( n − 5 ) = 0
Therefore, n = 15 n = 15 n = 15 or n = 5 n = 5 n = 5 .
Conclusion: The terms in the quadratic sequence that have a value of 56 are for n = 5 n = 5 n = 5 and n = 15 n = 15 n = 15 .
Hence, or otherwise, calculate the value of \( \sum_{n=5}^{10} T_n - \sum_{n=1}^{15} T_n \). Only available for registered users.
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To calculate the summation,
First, identify the terms:
Using symmetry, since T 10 = T 15 T_{10} = T_{15} T 10 = T 15 and T 5 = T 10 T_{5} = T_{10} T 5 = T 10 ,
We find the terms:
For n = 5 : T 5 = 56 n=5: T_5 = 56 n = 5 : T 5 = 56 ,
For n = 10 : T 10 = 81 n=10: T_{10} = 81 n = 10 : T 10 = 81 ,
For n = 15 : T 15 = 56 n=15: T_{15} = 56 n = 15 : T 15 = 56 .
Then, compute the summations:
∑ n = 5 10 T n = T 5 + T 6 + T 7 + T 8 + T 9 + T 10 \sum_{n=5}^{10} T_n = T_5 + T_6 + T_7 + T_8 + T_9 + T_{10} ∑ n = 5 10 T n = T 5 + T 6 + T 7 + T 8 + T 9 + T 10
The respective terms will sum to:
T 5 = 56 T_5 = 56 T 5 = 56 ,
T 6 = 72 T_6 = 72 T 6 = 72 ,
T 7 = 80 T_7 = 80 T 7 = 80 ,
T 8 = 81 T_8 = 81 T 8 = 81 ,
T 9 = 80 T_9 = 80 T 9 = 80 ,
T 10 = 81 T_{10} = 81 T 10 = 81 .
Total: 56 + 72 + 80 + 81 + 80 + 81 = 450 56 + 72 + 80 + 81 + 80 + 81 = 450 56 + 72 + 80 + 81 + 80 + 81 = 450 .
Calculating (\sum_{n=1}^{15} T_n ):
This will yield the total result of 81 using previous evaluations.
Finally, subtract the two sums:
450 − 81 = 369 450 - 81 = 369 450 − 81 = 369
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