For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Solved Examples on Quadratic Sequences

Samuel Dominic Chukwuemeka (SamDom For Peace) Prerequisites:
(1.) Linear Systems
(2.) Quadratic Equations
(3.) Quadratic Functions

Formulas: Formulas
Calculators: Calculators


For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Where applicable, the Formulas are used as a second method.
Please ask your teacher if it will be allowed if you memorize it well.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
Where applicable, the Formulas are used as a second method.
Please ask your teacher if it will be allowed if you memorize it well
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
Use at least two methods whenever applicable.
Show all work

(1.) GCSE Here are the first four terms of a quadratic sequence.

$11\:\:\:\:\:26\:\:\:\:\:45\:\:\:\:\:68 \\[3ex]$ Work out an expression for the $nth$ term


$ QS_n = an^2 + bn + c \\[3ex] \underline{First\:\:Method} \\[3ex] 1st\:\:term = 11 \\[3ex] 1st\:\:term \rightarrow n = 1 \\[3ex] QS_1 = a(1)^2 + b(1) + c = 11 \\[3ex] a + b + c = 11...eqn.(1) \\[3ex] 2nd\:\:term = 26 \\[3ex] 2nd\:\:term \rightarrow n = 2 \\[3ex] QS_2 = a(2)^2 + b(2) + c = 26 \\[3ex] a(4) + 2b + c = 26 \\[3ex] 4a + 2b + c = 26...eqn.(2) \\[3ex] 3rd\:\:term = 45 \\[3ex] 3rd\:\:term \rightarrow n = 3 \\[3ex] QS_3 = a(3)^2 + b(3) + c = 45 \\[3ex] a(9) + 3b + c = 45 \\[3ex] 9a + 3b + c = 45...eqn.(3) \\[3ex] eqn.(2) - eqn.(1) \rightarrow 3a + b = 15...eqn.(4) \\[3ex] eqn.(3) - eqn.(1) \rightarrow 8a + 2b = 34...eqn.(5) \\[3ex] 2 * eqn.(4) = 6a + 2b = 30...eqn.(6) \\[3ex] eqn.(5) - eqn.(6) \rightarrow 2a = 4 \\[3ex] a = \dfrac{4}{2} \\[5ex] a = 2 \\[3ex] From\:\:eqn.(4);\:\: b = 15 - 3a \\[3ex] b = 15 - 3(2) \\[3ex] b = 15 - 6 \\[3ex] b = 9 \\[3ex] From\:\:eqn.(1);\:\: c = 11 - a - b \\[3ex] c = 11 - 2 - 9 \\[3ex] c = 0 \\[3ex] QS_n = 2n^2 + 9n + 0 \\[3ex] \therefore QS_n = 2n^2 + 9n \\[3ex] \underline{Second\:\:Method:\:\:By\:\:Formula} \\[3ex] 1st = 11 \\[3ex] 2nd = 26 \\[3ex] 3rd = 45 \\[3ex] a = \dfrac{1st + 3rd - 2(2nd)}{2} \\[5ex] a = \dfrac{11 + 45 - 2(26)}{2} \\[5ex] = \dfrac{11 + 45 - 52}{2} \\[5ex] = \dfrac{56 - 52}{2} \\[5ex] = \dfrac{4}{2} \\[5ex] a = 2 \\[3ex] b = \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} \\[5ex] b = \dfrac{8(26) - 5(11) - 3(45)}{2} \\[5ex] = \dfrac{208 - 55 - 135}{2} \\[5ex] = \dfrac{18}{2} \\[5ex] b = 9 \\[3ex] c = 3(1st) - 3(2nd) + 3rd \\[3ex] c = 3(11) - 3(26) + 45 \\[3ex] c = 33 - 78 + 45 \\[3ex] c = 0 \\[3ex] QS_n = 2n^2 + 9n + 0 \\[3ex] \therefore QS_n = 2n^2 + 9n $
(2.) GCSE Work out the next term of this quadratic sequence.

$5\:\:\:\:\:8\:\:\:\:\:14\:\:\:\:\:23\:\:\:\:.... \\[3ex]$

$ Let\:\:the\:\:value\:\:of\:\:the\:\:next\:\:term = c \\[3ex] 5;\:\:\:8;\:\:\:14;\:\:\:23;\:\:\:c \\[3ex] First\:\:Difference: \\[3ex] 8 - 5 = 3 \\[3ex] 14 - 8 = 6 \\[3ex] 23 - 14 = 9 \\[3ex] c - 23 = c - 23 \\[3ex] Second\:\:Difference: \\[3ex] 6 - 3 = 3 \\[3ex] 9 - 6 = 3 \\[3ex] c - 23 - 9 = c - 32 \\[3ex] This\:\:should\:\:be\:\:equal\:\:to\:\: 3 \\[3ex] c - 32 = 3 \\[3ex] c = 3 + 32 \\[3ex] c = 35 \\[3ex] The\:\:next\:\:term = 35 $
(3.) GCSE Work out an expression for the $nth$ term of the quadratic sequence

$2\:\:\:\:\:17\:\:\:\:\:40\:\:\:\:\:71\:\:\:\:.... \\[3ex]$

$ QS_n = an^2 + bn + c \\[3ex] \underline{First\:\:Method} \\[3ex] 1st\:\:term = 2 \\[3ex] 1st\:\:term \rightarrow n = 1 \\[3ex] QS_1 = a(1)^2 + b(1) + c = 2 \\[3ex] a + b + c = 2...eqn.(1) \\[3ex] 2nd\:\:term = 17 \\[3ex] 2nd\:\:term \rightarrow n = 2 \\[3ex] QS_2 = a(2)^2 + b(2) + c = 17 \\[3ex] a(4) + 2b + c = 17 \\[3ex] 4a + 2b + c = 17...eqn.(2) \\[3ex] 3rd\:\:term = 40 \\[3ex] 3rd\:\:term \rightarrow n = 3 \\[3ex] QS_3 = a(3)^2 + b(3) + c = 40 \\[3ex] a(9) + 3b + c = 40 \\[3ex] 9a + 3b + c = 40...eqn.(3) \\[3ex] eqn.(2) - eqn.(1) \rightarrow 3a + b = 15...eqn.(4) \\[3ex] eqn.(3) - eqn.(1) \rightarrow 8a + 2b = 38...eqn.(5) \\[3ex] 2 * eqn.(4) = 6a + 2b = 30...eqn.(6) \\[3ex] eqn.(5) - eqn.(6) \rightarrow 2a = 8 - 2 \\[3ex] 2a = 8 \\[3ex] a = \dfrac{8}{2} \\[5ex] a = 4 \\[3ex] From\:\:eqn.(4);\:\: b = 15 - 3a \\[3ex] b = 15 - 3(4) \\[3ex] b = 15 - 12 \\[3ex] b = 3 \\[3ex] From\:\:eqn.(1);\:\: c = 2 - a - b \\[3ex] c = 2 - 4 - 3 \\[3ex] c = -5 \\[3ex] QS_n = 4n^2 + 3n + -5 \\[3ex] \therefore QS_n = 4n^2 + 3n - 5 \\[3ex] \underline{Second\:\:Method:\:\:By\:\:Formula} \\[3ex] 1st = 2 \\[3ex] 2nd = 17 \\[3ex] 3rd = 40 \\[3ex] a = \dfrac{1st + 3rd - 2(2nd)}{2} \\[5ex] a = \dfrac{2 + 40 - 2(17)}{2} \\[5ex] = \dfrac{2 + 40 - 34}{2} \\[5ex] = \dfrac{8}{2} \\[5ex] a = 4 \\[3ex] b = \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} \\[5ex] b = \dfrac{8(17) - 5(2) - 3(40)}{2} \\[5ex] = \dfrac{136 - 10 - 120}{2} \\[5ex] = \dfrac{6}{2} \\[5ex] b = 3 \\[3ex] c = 3(1st) - 3(2nd) + 3rd \\[3ex] c = 3(2) - 3(17) + 40 \\[3ex] c = 6 - 51 + 40 \\[3ex] c = -5 \\[3ex] QS_n = 4n^2 + 3n + -5 \\[3ex] \therefore QS_n = 4n^2 + 3n - 5 \\[3ex] The\:\:n^{th}\:\:term\:\:of\:\:the\:\:sequence = 4n^2 + 3n - 5 $
(4.) NSC Given the quadratic sequence: $2;\:\:\:3;\:\:\:10;\:\:\:23;...$
(4.1.1) Write down the next term of the sequence.
(4.1.2) Determine the $n^{th}$ term of the sequence.
(4.1.3) Calculate the $20^{th}$ term of the sequence.


For a Quadratic Sequence, the Second Difference must be the same

$ (4.1.1) \\[3ex] Let\:\:the\:\:value\:\:of\:\:the\:\:next\:\:term = c \\[3ex] 2;\:\:\:3;\:\:\:10;\:\:\:23;\:\:\:c \\[3ex] First\:\:Difference: \\[3ex] 3 - 2 = 1 \\[3ex] 10 - 3 = 7 \\[3ex] 23 - 10 = 13 \\[3ex] c - 23 = c - 23 \\[3ex] Second\:\:Difference: \\[3ex] 7 - 1 = 6 \\[3ex] 13 - 7 = 6 \\[3ex] c - 23 - 13 = c - 36 \\[3ex] This\:\:should\:\:be\:\:equal\:\:to\:\: 6 \\[3ex] c - 36 = 6 \\[3ex] c = 6 + 36 \\[3ex] c = 42 \\[3ex] The\:\:next\:\:term = 42 \\[3ex] (4.1.2) \\[3ex] QS_n = an^2 + bn + c \\[3ex] \underline{First\:\:Method} \\[3ex] 1st\:\:term = 2 \\[3ex] 1st\:\:term \rightarrow n = 1 \\[3ex] QS_1 = a(1)^2 + b(1) + c = 2 \\[3ex] a + b + c = 2...eqn.(1) \\[3ex] 2nd\:\:term = 3 \\[3ex] 2nd\:\:term \rightarrow n = 2 \\[3ex] QS_2 = a(2)^2 + b(2) + c = 3 \\[3ex] a(4) + 2b + c = 3 \\[3ex] 4a + 2b + c = 3...eqn.(2) \\[3ex] 3rd\:\:term = 10 \\[3ex] 3rd\:\:term \rightarrow n = 3 \\[3ex] QS_3 = a(3)^2 + b(3) + c = 10 \\[3ex] a(9) + 3b + c = 10 \\[3ex] 9a + 3b + c = 10...eqn.(3) \\[3ex] eqn.(2) - eqn.(1) \rightarrow 3a + b = 1...eqn.(4) \\[3ex] eqn.(3) - eqn.(1) \rightarrow 8a + 2b = 8...eqn.(5) \\[3ex] 2 * eqn.(4) = 6a + 2b = 2...eqn.(6) \\[3ex] eqn.(5) - eqn.(6) \rightarrow 2a = 8 - 2 \\[3ex] 2a = 6 \\[3ex] a = \dfrac{6}{2} \\[5ex] a = 3 \\[3ex] From\:\:eqn.(4);\:\: b = 1 - 3a \\[3ex] b = 1 - 3(3) \\[3ex] b = 1 - 9 \\[3ex] b = -8 \\[3ex] From\:\:eqn.(1);\:\: c = 2 - a - b \\[3ex] c = 2 - 3 - (-8) \\[3ex] c = -1 + 8 \\[3ex] c = 7 \\[3ex] QS_n = 3n^2 + -8n + 7 \\[3ex] \therefore QS_n = 3n^2 - 8n + 7 \\[3ex] \underline{Second\:\:Method:\:\:By\:\:Formula} \\[3ex] 1st = 2 \\[3ex] 2nd = 3 \\[3ex] 3rd = 10 \\[3ex] a = \dfrac{1st + 3rd - 2(2nd)}{2} \\[5ex] a = \dfrac{2 + 10 - 2(3)}{2} \\[5ex] = \dfrac{2 + 10 - 6}{2} \\[5ex] = \dfrac{6}{2} \\[5ex] a = 3 \\[3ex] b = \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} \\[5ex] b = \dfrac{8(3) - 5(2) - 3(10)}{2} \\[5ex] = \dfrac{24 - 10 - 30}{2} \\[5ex] = \dfrac{-16}{2} \\[5ex] b = -8 \\[3ex] c = 3(1st) - 3(2nd) + 3rd \\[3ex] c = 3(2) - 3(3) + 10 \\[3ex] c = 6 - 9 + 10 \\[3ex] c = 7 \\[3ex] QS_n = 3n^2 + -8n + 7 \\[3ex] \therefore QS_n = 3n^2 - 8n + 7 \\[3ex] The\:\:n^{th}\:\:term\:\:of\:\:the\:\:sequence = 3n^2 - 8n + 7 \\[3ex] (2.1.3) \\[3ex] QS_n = 3n^2 - 8n + 7 \\[3ex] For\:\: n = 20 \\[3ex] QS_{20} = 3 * (20)^2 - 8(20) + 7 \\[3ex] = 3 * 400 - 160 + 7 \\[3ex] = 1200 - 160 + 7 \\[3ex] QS_{20} = 1047 \\[3ex] The\:\:20^{th}\:\:term\:\:of\:\:the\:\:sequence = 1047 $
(5.) NSC Given the quadratic sequence: $321;\:\:\:290;\:\:\:261;\:\:\:234;...$
(5.1.1) Write down the values of the next TWO terms of the sequence.
(5.1.2) Determine the general term of the sequence in the form $T_n = an^2 + bn + c$.
(5.1.3) Which term(s) of the sequence will have a value of $74$?
(5.1.4) Which term in the sequence has the least value?


For a Quadratic Sequence, the Second Difference must be the same

$ (5.1.1) \\[3ex] Let\:\:the\:\:values\:\:of\:\:the\:\:next\:\:two\:\:terms = c\:\:and\:\:d \\[3ex] 321;\:\:\:290;\:\:\:261;\:\:\:234;\:\:\:c;\:\:\:d \\[3ex] First\:\:Difference: \\[3ex] 290 - 321 = -31 \\[3ex] 261 - 290 = -29 \\[3ex] 234 - 261 = -27 \\[3ex] c - 234 = c - 234 \\[3ex] d - c = d - c \\[3ex] Second\:\:Difference: \\[3ex] -29 - (-31) = -29 + 31 = 2 \\[3ex] -27 - (-29) = -27 + 29 = 2 \\[3ex] c - 234 - (-27) = c - 234 + 27 = c - 207 \\[3ex] This\:\:should\:\:be\:\:equal\:\:to\:\: 2 \\[3ex] c - 207 = 2 \\[3ex] c = 2 + 207 \\[3ex] c = 209 \\[3ex] Also:\:\: d - c - (c - 234) = 2 \\[3ex] d - c - c + 234 = 2 d - 2c + 234 = 2 \\[3ex] d = 2 + 2c - 234 \\[3ex] d = 2 + 2(209) - 234 \\[3ex] d = 2 + 418 - 234 \\[3ex] d = 186 \\[3ex] The\:\:next\:\:two\:\:terms = 209;\:\:\:186 \\[3ex] (5.1.2) \\[3ex] QS_n = an^2 + bn + c \\[3ex] \underline{First\:\:Method} \\[3ex] 1st\:\:term = 321 \\[3ex] 1st\:\:term \rightarrow n = 1 \\[3ex] QS_1 = a(1)^2 + b(1) + c = 321 \\[3ex] a + b + c = 321...eqn.(1) \\[3ex] 2nd\:\:term = 290 \\[3ex] 2nd\:\:term \rightarrow n = 2 \\[3ex] QS_2 = a(2)^2 + b(2) + c = 290 \\[3ex] a(4) + 2b + c = 290 \\[3ex] 4a + 2b + c = 290...eqn.(2) \\[3ex] 3rd\:\:term = 261 \\[3ex] 3rd\:\:term \rightarrow n = 3 \\[3ex] QS_3 = a(3)^2 + b(3) + c = 261 \\[3ex] a(9) + 3b + c = 261 \\[3ex] 9a + 3b + c = 261...eqn.(3) \\[3ex] eqn.(2) - eqn.(1) \rightarrow 3a + b = -31...eqn.(4) \\[3ex] eqn.(3) - eqn.(1) \rightarrow 8a + 2b = -60...eqn.(5) \\[3ex] 2 * eqn.(4) = 6a + 2b = -62...eqn.(6) \\[3ex] eqn.(5) - eqn.(6) \rightarrow 2a = -60 - (-62) \\[3ex] 2a = -60 + 62 \\[3ex] 2a = 2 \\[3ex] a = \dfrac{2}{2} \\[5ex] a = 1 \\[3ex] From\:\:eqn.(4);\:\: b = -31 - 3a \\[3ex] b = -31 - 3(1) \\[3ex] b = -31 - 3 \\[3ex] b = -34 \\[3ex] From\:\:eqn.(1);\:\: c = 321 - a - b \\[3ex] c = 321 - 1 - (-34) \\[3ex] c = 320 + 34 \\[3ex] c = 354 \\[3ex] QS_n = 1n^2 + -34n + 354 \\[3ex] \therefore QS_n = n^2 - 34n + 354 \\[3ex] \underline{Second\:\:Method:\:\:By\:\:Formula} \\[3ex] 1st = 321 \\[3ex] 2nd = 290 \\[3ex] 3rd = 261 \\[3ex] a = \dfrac{1st + 3rd - 2(2nd)}{2} \\[5ex] a = \dfrac{321 + 261 - 2(290)}{2} \\[5ex] = \dfrac{321 + 261 - 580}{2} \\[5ex] = \dfrac{2}{2} \\[5ex] a = 1 \\[3ex] b = \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} \\[5ex] b = \dfrac{8(290) - 5(321) - 3(261)}{2} \\[5ex] = \dfrac{2320 - 1605 - 783}{2} \\[5ex] = \dfrac{-68}{2} \\[5ex] b = -34 \\[3ex] c = 3(1st) - 3(2nd) + 3rd \\[3ex] c = 3(321) - 3(290) + 261 \\[3ex] c = 963 - 870 + 261 \\[3ex] c = 354 \\[3ex] QS_n = 1n^2 + -34n + 354 \\[3ex] \therefore QS_n = n^2 - 34n + 354 \\[3ex] (5.1.3) \\[3ex] n^2 - 34n + 354 = 74 \\[3ex] n^2 - 34n + 354 - 74 = 0 \\[3ex] n^2 - 34n + 280 = 0 \\[3ex] n^2 - 14n - 20n + 280 = 0 \\[3ex] n^2 - 14n = n(n - 14) \\[3ex] -20n + 280 = -20(n - 14) \\[3ex] (n - 14)(n - 20) = 0 \\[3ex] n - 14 = 0 \:\:\:OR\:\:\: n - 20 = 0 \\[3ex] n = 14 \:\:\:OR\:\:\: n = 20 \\[3ex] \underline{Check} \\[3ex] QS_n = n^2 - 34n + 354 \\[3ex] For\:\: n = 14 \\[3ex] QS_{14} = (14)^2 - 34(14) + 354 \\[3ex] QS_{14} = 196 - 476 + 354 \\[3ex] QS_{14} = 74 \checkmark \\[3ex] For\:\: n = 20 \\[3ex] QS_{20} = (20)^2 - 34(20) + 354 \\[3ex] QS_{20} = 400 - 680 + 354 \\[3ex] QS_{20} = 74 \checkmark \\[3ex] (5.1.4) \\[3ex] Least\:\:value\:\:\implies Vertex\:\:of\:\:the\:\:Quadratic\:\:Function \\[3ex] x-coordinate\:\:of\:\:the\:\:Vertex = term\:\:that\:\:has\:\:the\:\:least\:\:value \\[3ex] y-coordinate\:\:of\:\:the\:\:Vertex = least\:\:value \\[3ex] QS_n = n^2 - 34n + 354 \\[3ex] a = 1 \\[3ex] b = -34 \\[3ex] x = -\dfrac{b}{2a} \\[5ex] x = -\dfrac{-34}{2(1)} \\[5ex] x = \dfrac{34}{2} \\[5ex] x = 17 \\[3ex] The\:\:17th\:\:term\:\:has\:\:the\:\:least\:\:value \\[3ex] \underline{Extra} \\[3ex] Find\:\:the\:\:least\:\:value\:\:in\:\:the\:\:sequence \\[3ex] y-coordinate\:\:of\:\:the\:\:Vertex = least\:\:value \\[3ex] QS_n = n^2 - 34n + 354 \\[3ex] For\:\: n = 17 \\[3ex] QS_{17} = (17)^2 - 34(17) + 354 \\[3ex] QS_{17} = 289 - 578 + 354 \\[3ex] QS_{17} = 65 $
(6.) JAMB The $nth$ term of a sequence is $n^2 - 6n - 4$.
Find the sum of the $3rd$ and the $4th$ terms.

$ A.\:\: 24 \\[3ex] B.\:\: 23 \\[3ex] C.\:\: -24 \\[3ex] D.\:\: -25 \\[3ex] $

$ QS_n = n^2 - 6n - 4 \\[3ex] QS_3 = 3^2 - 6(3) - 4 \\[3ex] = 9 - 18 - 4 \\[3ex] = -13 \\[3ex] QS_4 = 4^2 - 6(4) - 4 \\[3ex] = 16 - 24 - 4 \\[3ex] = -12 \\[3ex] QS_3 + QS_4 \\[3ex] = -13 + (-12) \\[3ex] = -13 - 12 \\[3ex] = -25 $
(7.)


(8.)