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 All Sequences (Combination of Sequences)

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

Formulas: Formulas
Calculators: Calculators


For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT 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.
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.) NSC For which value of $n$ will the sum of the first $n$ terms of the arithmetic sequence in QUESTION $(14.)$ of Arithmetic Sequences be equal to the $n^{th}$ term of the Quadratic Sequence in QUESTION $(4.)$


$ \underline{Arithmetic\:\:Sequence} \\[3ex] AS_n = a + d(n - 1) \\[3ex] a = 35 \\[3ex] d = -7 \\[3ex] SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] SAS_n = \dfrac{n}{2}[2(35) + -7(n - 1)] \\[5ex] = \dfrac{n}{2}[70 - 7(n - 1)] \\[5ex] = \dfrac{n}{2}[70 - 7n + 7] \\[5ex] = \dfrac{n}{2}[77 - 7n] \\[5ex] \underline{Quadratic\:\:Sequence} \\[3ex] QS_n = 3n^2 - 8n + 7 \\[3ex] SAS_n = QS_n \\[3ex] \dfrac{n}{2}[77 - 7n] = 3n^2 - 8n + 7 \\[5ex] LCD = 2 \\[3ex] 2 * \dfrac{n}{2}[77 - 7n] = 2(3n^2 - 8n + 7) \\[5ex] n(77 - 7n) = 6n^2 - 16n + 14 \\[3ex] 77n - 7n^2 = 6n^2 - 16n + 14 \\[3ex] 6n^2 - 16n + 14 = 77n - 7n^2 \\[3ex] 6n^2 + 7n^2 - 16n - 77n + 14 = 0 \\[3ex] 13n^2 - 93n + 14 = 0 \\[3ex] 13n^2 - 91n - 2n + 14 = 0 \\[3ex] 13n^2 - 91n = 13n(n - 7) \\[3ex] -2n + 14 = -2(n - 7) \\[3ex] (n - 7)(13n - 2) = 0 \\[3ex] n - 7 = 0 \:\:\:OR\:\:\: 13n - 2 = 0 \\[3ex] n = 7 \:\:\:OR\:\:\: 13n = 2 \\[3ex] n = 7 \:\:\:OR\:\:\: n = \dfrac{2}{13} \\[5ex] n\:\:cannot\:\:be\:\:a\:\:fraction \\[3ex] n = 7 $
(2.) ACT A sequence of $5$ numbers has $6$ as its first term and $32$ as its last term.
The first $3$ numbers are an arithmetic sequence.
The last $3$ numbers are a geometric sequence with a common ratio of $2$.
What is the common difference among the first $3$ terms?

$ F.\:\: 0 \\[3ex] G.\:\: 1 \\[3ex] H.\:\: 61 \\[3ex] J.\:\: 67 \\[3ex] K.\:\: 72 \\[3ex] $

Let us set up the sequence...have a visual representation

$ .....~~~~.....~~~~.....~~~~.....~~~~.....~~~~..... \\[3ex] |~~~~~~~~~~~~~~~~AS~~~~~~~~~~~~~~~~|~~~~~~~~~~~~~~GS~~~~~~~~~~~~~~| \\[3ex] For\:\:the\:\:AS: \\[3ex] First\:\:term = a = 6 \\[3ex] Second\:\:term = a + d = 6 + d \\[3ex] Third\:\:term = a + 2d = 6 + 2d \\[3ex] For the GS: \\[3ex] Fourth\:\:term = r * (6 + 2d) = r(6 + 2d) \\[3ex] Fifth\:\:term = r * r * (6 + 2d) = r^2(6 + 2d) \\[3ex] Fifth\:\:term\:\:is\:\:also\:\:32 \\[3ex] 6,~~~~6 + d,~~~~6 + 2d,~~~~r(6 + d),~~~~r^2(6 + d) \\[3ex] 6,~~~~6 + d,~~~~6 + 2d,~~~~r(6 + d),~~~~32 \\[3ex] \rightarrow r^2(6 + 2d) = 32 \\[3ex] r = 2...given \\[3ex] 6 + 2d = \dfrac{32}{r^2} \\[5ex] 6 + 2d = \dfrac{32}{2^2} \\[5ex] 6 + 2d = \dfrac{32}{4} \\[5ex] 6 + 2d = 8 \\[3ex] 2d = 8 - 6 \\[3ex] 2d = 2 \\[3ex] d = \dfrac{2}{2} \\[5ex] d = 1...correct\:\:answer \\[3ex] Explore\:\:more \\[3ex] \underline{Check} \\[3ex] 6 + d = 6 + 1 = 7 \\[3ex] 6 + 2d = 6 + 2(1) = 6 + 2 = 8 \\[3ex] r(6 + d) = 2(7) = 14 \\[3ex] r^2(6 + d) = 32...given \\[3ex] Sequence\:\:is\:\: 6, 7, 8, 16, 32 \\[3ex] 6, 7, 8\:\:is\:\:an\:\:AS \\[3ex] 8, 16, 32\:\:is\:\:a\:\:GS $
(3.) WASSCE-FM The sum of the first $n$ terms of a sequence is given by $S_n = \dfrac{5n^2}{2} + \dfrac{5n}{2}$

Write down the first four terms of the sequence and find an exprsssion for the $nth$ term.


We can solve this question in at least two ways.
Use any method you prefer.

We are given the sum of the first $n$ terms
However, we do not know the kind of sequence
So, let us try to determine the kind of sequence by testing some numbers
Let us find the sum of the first one term; sum of the first two terms; sum of the first three terms; sum of the first four terms
Then, we can find the four terms of the sequence
And determine the fifth term and the sum of the first five terms
And verify that sum by finding the sum of the first five terms.

$ \underline{First\:\:Method} \\[3ex] Let\:\:the\:\:terms\:\:be\:\:a, b, c, d, e \\[3ex] S_1 = sum\:\:of\:\:the\:\:first\:\:1\:\:term \\[3ex] S_2 = sum\:\:of\:\:the\:\:first\:\:2\:\:terms \\[3ex] S_3 = sum\:\:of\:\:the\:\:first\:\:3\:\:terms \\[3ex] S_4 = sum\:\:of\:\:the\:\:first\:\:4\:\:terms \\[3ex] S_5 = sum\:\:of\:\:the\:\:first\:\:5\:\:terms \\[3ex] S_n = \dfrac{5n^2}{2} + \dfrac{5n}{2} \\[5ex] S_1 = \dfrac{5(1)^2}{2} + \dfrac{5(1)}{2} \\[5ex] = \dfrac{5(1)}{2} + \dfrac{5}{2} \\[5ex] = \dfrac{5}{2} + \dfrac{5}{2} \\[5ex] = \dfrac{5 + 5}{2} \\[5ex] = \dfrac{10}{2} \\[5ex] S_1 = 5 \\[3ex] \rightarrow a = 5 \\[3ex] S_2 = \dfrac{5(2)^2}{2} + \dfrac{5(2)}{2} \\[5ex] = \dfrac{5(4)}{2} + \dfrac{10}{2} \\[5ex] = \dfrac{20}{2} + 5 \\[5ex] = 10 + 5 \\[3ex] S_2 = 15 \\[3ex] \rightarrow a + b = 15 \\[3ex] 5 + b = 15 \\[3ex] b = 15 - 5 \\[3ex] b = 10 \\[3ex] S_3 = \dfrac{5(3)^2}{2} + \dfrac{5(3)}{2} \\[5ex] = \dfrac{5(9)}{2} + \dfrac{15}{2} \\[5ex] = \dfrac{45}{2} + \dfrac{15}{2} \\[5ex] = \dfrac{45 + 15}{2} \\[5ex] = \dfrac{60}{2} \\[5ex] S_3 = 30 \\[3ex] \rightarrow a + b + c = 30 \\[3ex] 15 + c = 30 \\[3ex] c = 30 - 15 \\[3ex] c = 15 \\[3ex] a, b, c = 5, 10, 15...is\:\:an\:\:arithmetic\:\:sequence \\[3ex] But,\:\:we\:\:can\:\:still\:\:explore\:\:more \\[3ex] S_4 = \dfrac{5(4)^2}{2} + \dfrac{5(4)}{2} \\[5ex] = \dfrac{5(16)}{2} + \dfrac{20}{2} \\[5ex] = 5(8) + 10 \\[3ex] = 40 + 10 \\[3ex] S_4 = 50 \\[3ex] \rightarrow a + b + c + d = 50 \\[3ex] 30 + d = 50 \\[3ex] d = 50 - 30 \\[3ex] d = 20 \\[3ex] a, b, c, d = 5, 10, 15, 20 \\[3ex] a, b, c, d, e = 5, 10, 15, 20, 25 \\[3ex] a + b + c + d + e = 5 + 10 + 15 + 20 + 25 = 75 \\[3ex] Verify \\[3ex] S_5 = \dfrac{5(5)^2}{2} + \dfrac{5(5)}{2} \\[5ex] = \dfrac{5(25)}{2} + \dfrac{25}{2} \\[5ex] = \dfrac{125}{2} + \dfrac{25}{2} \\[5ex] = \dfrac{125 + 25}{2} \\[5ex] = \dfrac{150}{2} \\[5ex] S_5 = 75...verified \\[3ex] \therefore AP:\:\: 5, 10, 15, 20 \\[3ex] a = 5 \\[3ex] d = 10 - 5 = 5 \\[3ex] AP_n = a + d(n - 1) \\[3ex] AP_n = 5 + 5(n - 1) \\[3ex] AP_n = 5 + 5n - 5 \\[3ex] AP_n = 5n \\[3ex] \underline{Second\:\:Method} \\[3ex] S_n = \dfrac{5n^2}{2} + \dfrac{5n}{2} \\[5ex] S_n = \dfrac{n}{2}(5n + 5) \\[5ex] S_n = \dfrac{n}{2}(5 + 5n) \\[5ex] Compare\:\:to\:\:the\:\:Sum\:\:of\:\:the\:\:first\:\:n\:\:terms\:\:of\:\:an\:\:A.P \\[3ex] S_n = \dfrac{n}{2}(a + AP_n) \\[5ex] \rightarrow a = 5 \\[3ex] \rightarrow AP_n = 5n $
(4.) WASSCE-FM (a.) If $3, x, y, 18$ are in Arithmetic Progression (A.P), find the values of $x$ and $y$
(b.) (i.) The sum of the second and third terms of a geometric progression is six times the fourth term.
Find the two possible values of the common ratio.
(ii.) If the second term is $8$ and the common ratio is positive, find the first six terms.


$ (a.) \\[3ex] AP:\:\: 3, x, y, 18 \\[3ex] d = x - 3 \\[3ex] d = y - x \\[3ex] d = 18 - y \\[3ex] d = d \\[3ex] \rightarrow x - 3 = y - x \\[3ex] 2x - 3 = y \\[3ex] y = 2x - 3...eqn.(1) \\[3ex] \rightarrow y - x = 18 - y \\[3ex] 2y - x = 18 \\[3ex] Substitute\:\: (2x - 3)\:\:for\:\:y \\[3ex] 2(2x - 3) - x = 18 \\[3ex] 4x - 6 - x = 18 \\[3ex] 3x - 6 = 18 \\[3ex] 3x = 18 + 6 \\[3ex] 3x = 24 \\[3ex] x = \dfrac{24}{3} \\[5ex] x = 8 \\[3ex] Substitute\:\:x = 8\:\:in\:\:eqn.(1) \\[3ex] y = 2x - 3 \\[3ex] y = 2(8) - 3 \\[3ex] y = 16 - 3 \\[3ex] y = 13 \\[3ex] AP:\:\: 3, 8, 13, 18 \\[3ex] (b.) \\[3ex] (i.) \\[3ex] GP_n = ar^{n - 1} \\[3ex] GP_2 = ar \\[3ex] GP_3 = ar^2 \\[3ex] GP_4 = ar^3 \\[3ex] GP_2 + GP_3 = 6 * GP_4 \\[3ex] ar + ar^2 = 6(ar^3) \\[3ex] a(r + r^2) = 6ar^3 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:a \\[3ex] r + r^2 = 6r^3 \\[3ex] 0 = 6r^3 - r^2 - r \\[3ex] 6r^3 - r^2 - r = 0 \\[3ex] r(6r^2 - r - 1) = 0 \\[3ex] Factor\:\: 6r^2 - r - 1 \\[3ex] 6r^2 + 2r - 3r - 1 \\[3ex] 2r(3r + 1) - 1(3r + 1) \\[3ex] (3r + 1)(2r - 1) \\[3ex] \rightarrow r(3r + 1)(2r - 1) = 0 \\[3ex] r = 0 \:\:OR \\[3ex] 3r + 1 = 0 \\[3ex] 3r = 0 - 1 \\[3ex] 3r = -1 \\[3ex] r = -\dfrac{1}{3} \:\:OR \\[5ex] 2r - 1 = 0 \\[3ex] 2r = 0 + 1 \\[3ex] 2r = 1 \\[3ex] r = \dfrac{1}{2} \\[5ex] r = -\dfrac{1}{3},0,\dfrac{1}{2} \\[5ex] r \ne 0 \\[3ex] \therefore r = -\dfrac{1}{3},\dfrac{1}{2} \\[5ex] (ii.) \\[3ex] r \:\:is\:\:positive \\[3ex] \rightarrow r = \dfrac{1}{2} \\[5ex] GP_2 = ar = 8 \\[3ex] a = 8 \div r \\[3ex] a = 8 \div \dfrac{1}{2} \\[5ex] a = 8 * \dfrac{2}{1} \\[5ex] a = 16 \\[3ex] GP_3 = ar^2 = ar * r = GP_2 * r = 8 * \dfrac{1}{2} = 4 \\[5ex] GP_4 = GP_3 * r = 4 * \dfrac{1}{2} = 2 \\[5ex] GP_5 = GP_4 * r = 2 * \dfrac{1}{2} = 1 \\[5ex] GP_6 = GP_5 * r = 1 * \dfrac{1}{2} = \dfrac{1}{2} \\[5ex] GP:\:\: 16, 8, 4, 2, 1, \dfrac{1}{2} $
(5.)


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

$ (a.) \\[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] (b.) \\[3ex] $
(6.)


$ r = \dfrac{2nd\:\: term}{1st\:\: term} \\[5ex] r = \dfrac{abc^2d}{bcd} = \dfrac{a * b * c * c * d}{b * c * d} = ac \\[5ex] 4th\:\:term = 3rd\:\: term * r \\[3ex] 4th\:\:term = a^2bc^3d * ac = a^2 * a * b * c^3 * c * d = a^3bc^4d \\[3ex] 4th\:\:term = a^3bc^4d $
(7.)


(8.) ACT