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‘n Kwadraatiese ry het die volgende eienskappe: Die tweede verskil is 10: Die eerste twee terme is gelyk, d.w.s - NSC Mathematics - Question 3 - 2023 - Paper 1

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‘n-Kwadraatiese-ry-het-die-volgende-eienskappe:--Die-tweede-verskil-is-10:--Die-eerste-twee-terme-is-gelyk,-d.w.s-NSC Mathematics-Question 3-2023-Paper 1.png

‘n Kwadraatiese ry het die volgende eienskappe: Die tweede verskil is 10: Die eerste twee terme is gelyk, d.w.s. $T_1 = T_2$. $T_1 + T_2 + T_3 = 28$ 3.1 Toon dat... show full transcript

Worked Solution & Example Answer:‘n Kwadraatiese ry het die volgende eienskappe: Die tweede verskil is 10: Die eerste twee terme is gelyk, d.w.s - NSC Mathematics - Question 3 - 2023 - Paper 1

Step 1

3.1 Toon dat die algemene term van die ry $T_n = 5n^2 - 15n + 16$ is.

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Answer

To solve for the general term of the quadratic sequence, we start with the information given:

  1. The second difference is 10, which indicates that the coefficient of n2n^2 is half of the second difference. Thus, we have: 2a=10a=52a = 10 \Rightarrow a = 5

  2. The first two terms are equal, meaning: T1=T2T_1 = T_2

  3. The sum of the first three terms: T1+T2+T3=28T_1 + T_2 + T_3 = 28

Using the general form of a quadratic sequence:
Tn=an2+bn+cT_n = an^2 + bn + c Substituting a=5a = 5: Tn=5n2+bn+cT_n = 5n^2 + bn + c

We know:

  • T1=5(12)+b(1)+c=5+b+cT_1 = 5(1^2) + b(1) + c = 5 + b + c
  • T2=5(22)+b(2)+c=20+2b+cT_2 = 5(2^2) + b(2) + c = 20 + 2b + c

Since T1=T2T_1 = T_2, we can set the equations equal: 5+b+c=20+2b+c5 + b + c = 20 + 2b + c This simplifies to: b=15b = 15

  1. Plugging bb back into TnT_n gives us: Tn=5n215n+cT_n = 5n^2 - 15n + c Now substituting into the equation for the sum: T1+T2+T3=5+15+c=28T_1 + T_2 + T_3 = 5 + 15 + c = 28 Sufficient to determine: 20+c=28c=820 + c = 28 \Rightarrow c = 8

  2. Therefore, the general term is: Tn=5n215n+16T_n = 5n^2 - 15n + 16

Step 2

3.2 Is 216 ‘n term van hierdie ry? Motiveer jou antwoord met die nodige berekening.

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Answer

To determine if 216 is a term in the sequence given by: Tn=5n215n+16T_n = 5n^2 - 15n + 16 we set: Tn=216T_n = 216

This leads to the equation: 5n215n+16=2165n^2 - 15n + 16 = 216 Subtracting 216 from both sides gives: 5n215n200=05n^2 - 15n - 200 = 0

To simplify, we divide the entire equation by 5: n23n40=0n^2 - 3n - 40 = 0

Factoring the quadratic equation: (n8)(n+5)=0(n - 8)(n + 5) = 0 Thus: n=8orn=5n = 8 \quad \text{or} \quad n = -5

Since nn must be a positive integer, we conclude: n=8n = 8 Hence, 216 is indeed a term of the quadratic sequence corresponding to n=8n = 8.

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