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 Other Sequences

Samuel Dominic Chukwuemeka (SamDom For Peace) Prerequisites:
(1.) Linear Systems
(2.) Exponents
(3.) Quadratic Equations
(4.) 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
Indicate the kind of sequence as applicable.
Use at least two methods whenever applicable.
Show all work

(1.) WASSCE The $nth$ term of a sequence is $5 + \dfrac{2}{3^{n - 2}}$ for $n \ge 1$.
What is the sum of the fourth and fifth terms?
Leave your answers in the form $\dfrac{x}{y}$ where $x$ and $y$ are integers.


$ nth\:\:term = 5 + \dfrac{2}{3^{n - 2}} \\[5ex] 4th\:\:term = 5 + \dfrac{2}{3^{4 - 2}} \\[5ex] = 5 + \dfrac{2}{3^2} \\[5ex] = 5 + \dfrac{2}{9} \\[5ex] = \dfrac{45}{9} + \dfrac{2}{9} \\[5ex] = \dfrac{45 + 2}{9} \\[5ex] = \dfrac{47}{9} \\[5ex] 5th\:\:term = 5 + \dfrac{2}{3^{5 - 2}} \\[5ex] = 5 + \dfrac{2}{3^3} \\[5ex] = 5 + \dfrac{2}{27} \\[5ex] = \dfrac{135}{27} + \dfrac{2}{27} \\[5ex] = \dfrac{135 + 2}{27} \\[5ex] = \dfrac{137}{27} \\[5ex] 4th\:\:term + 5th\:\:term \\[3ex] = \dfrac{47}{9} + \dfrac{137}{27} \\[5ex] = \dfrac{141}{27} + \dfrac{137}{27} \\[5ex] = \dfrac{141 + 137}{27} \\[5ex] = \dfrac{278}{27} $
(2.) GCSE The term-to-term rule of a sequence is
Add $8$ and divide by $2$
The first term of the sequence is $-24$
Write out the next two terms.


$ 1st\:\:term = -24 \\[3ex] 2nd\:\:term = \dfrac{-24 + 8}{2} \\[5ex] = -\dfrac{16}{2} \\[5ex] = -8 \\[3ex] 3rd\:\:term = \dfrac{-8 + 8}{2} \\[5ex] = \dfrac{0}{2} \\[5ex] = 0 \\[3ex] $ The next two terms are $-8$ and $0$
(3.) ACT The sum of a sequence of consecutive odd numbers, where the smallest term is $1$, is always a perfect square.
For example, $1 + 3 = 2^2$ and $1 + 3 + 5 + 7 = 4^2$
One of the sequences described above has a sum of $144$
What is the largest odd number in the sequence?

$ F.\:\: 11 \\[3ex] G.\:\: 13 \\[3ex] H.\:\: 15 \\[3ex] J.\:\: 23 \\[3ex] K.\:\: 73 \\[3ex] $

Notice the pattern

$ 1 + 3\:\:are\:\:2\:\:numbers \\[3ex] 1 + 3 = 4 = 2^2 \\[3ex] 1 + 3 + 5\:\:are\:\:3\:\:numbers \\[3ex] 1 + 3 + 5 = 9 = 3^3 \\[3ex] 1 + 3 + 5 + 7 \:\:are\:\:4\:\:numbers \\[3ex] 1 + 3 + 5 + 7 = 16 = 4^2 \\[3ex] 1 + 3 + 5 + 7 + 9 \:\:are\:\:5\:\:numbers \\[3ex] 1 + 3 + 5 + 7 + 9 = 5^2 \\[3ex] Similarly \\[3ex] For\:\:144 \\[3ex] 144 = 12^2 \\[3ex] We\:\:need\:\:12\:\:numbers \\[3ex] 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 \:\:are\:\:12\:\:numbers \\[3ex] 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12^2 \\[3ex] Highest\:\:odd\:\:number = 23 $
(4.) ACT A sequence is defined for all positive integers by $s_n = 2s_{(n - 1)} + n + 1$ and $s_1 = 3$
What is $s_4?$

$ F.\:\: 9 \\[3ex] G.\:\: 18 \\[3ex] H.\:\: 22 \\[3ex] J.\:\: 49 \\[3ex] K.\:\: 111 \\[3ex] $

Recursive Seqeunce

$ s_n = 2s_{(n - 1)} + n + 1 \\[3ex] s_1 = 3 \\[3ex] s_2 = 2s_{(2 - 1)} + 2 + 1 \\[3ex] s_2 = 2s_1 + 2 + 1 \\[3ex] s_2 = 2(3) + 2 + 1 \\[3ex] s_2 = 6 + 2 + 1 \\[3ex] s_2 = 9 \\[3ex] s_3 = 2s_{(3 - 1)} + 3 + 1 \\[3ex] s_3 = 2s_2 + 3 + 1 \\[3ex] s_3 = 2(9) + 3 + 1 \\[3ex] s_3 = 18 + 3 + 1 \\[3ex] s_3 = 22 \\[3ex] s_4 = 2s_{(4 - 1)} + 4 + 1 \\[3ex] s_4 = 2s_3 + 4 + 1 \\[3ex] s_4 = 2(22) + 4 + 1 \\[3ex] s_4 = 44 + 4 + 1 \\[3ex] s_4 = 49 $
(5.) GCSE The term-to-term rule of a different sequence is
Subtract $1$ and multiply by $5$
The third term of this sequence is $120$
$........\:\:\:........\:\:\: 120$
Work out the first term.


$ 1st\:\:term = a \\[3ex] 2nd\:\:term = (a - 1) * 5 \\[3ex] = 5(a - 1) \\[3ex] = 5a - 5 \\[3ex] 3rd\:\:term = [(5a - 5) - 1] * 5 \\[3ex] = [5a - 5 - 1] * 5 \\[3ex] = (5a - 6) * 5 \\[3ex] = 5(5a - 6) \\[3ex] = 25a - 30 \\[3ex] 3rd\:\:term = 120 \\[3ex] \rightarrow 25a - 30 = 120 \\[3ex] 25a = 120 + 30 \\[3ex] 25a = 150 \\[3ex] a = \dfrac{150}{25} \\[5ex] a = 6 \\[3ex] $ The first term is $6$
(6.) GCSE Here is a rule for a sequence.
After the first two terms, each term is half the sum of the previous two terms.
Here is a sequence that follows this rule.
$2\:\:\:\:\:\:\:10\:\:\:\:\:\:\:6\:\:\:\:\:\:\:.......\:\:\:\:\:\:\:.......\:\:\:\:\:\:\:.......$
Show that the $6th$ term is the first one that is not a whole number.


$ 2\:\:\:\:\:\:\:10\:\:\:\:\:\:\:6\:\:\:\:\:\:\:.......\:\:\:\:\:\:\:.......\:\:\:\:\:\:\:....... \\[3ex] 1st\:\:term = 2 \\[3ex] 2nd\:\:term = 10 \\[3ex] 3rd\:\:term = 6 \\[3ex] 4th\:\:term = \dfrac{6 + 10}{2} = \dfrac{16}{2} = 8 \\[5ex] 5th\:\:term = \dfrac{8 + 6}{2} = \dfrac{14}{2} = 7 \\[5ex] 6th\:\:term = \dfrac{7 + 8}{2} = \dfrac{15}{2} = 7.5 \\[5ex] 7.5\:\:is\:\:not\:\:a\:\:whole\:\:number $
(7.) GCSE A sequence of numbers is formed by the iterative process

$ u_{n + 1} = \dfrac{3}{u_n + 1},\:\:\:\:\:\:\: u_1 = 4 \\[5ex] $ Work out the values of $u_2$ and $u_3$


Recursive Seqeunce

$ u_1 = 4 \\[3ex] u_{n + 1} = \dfrac{3}{u_n + 1} \\[5ex] For\:\: n = 1 \\[3ex] u_{1 + 1} = \dfrac{3}{u_1 + 1} \\[5ex] u_2 = \dfrac{3}{4 + 1} \\[5ex] u_2 = \dfrac{3}{5} \\[5ex] For\:\: n = 2 \\[3ex] u_{2 + 1} = \dfrac{3}{u_2 + 1} \\[5ex] u_3 = \dfrac{3}{\dfrac{3}{5} + 1} \\[7ex] \dfrac{3}{5} + 1 = \dfrac{3}{5} + \dfrac{5}{5} = \dfrac{3 + 5}{5} = \dfrac{8}{5} \\[5ex] u_3 = \dfrac{3}{\dfrac{8}{5}} \\[7ex] = 3 \div \dfrac{8}{5} \\[5ex] = 3 * \dfrac{5}{8} \\[5ex] u_3 = \dfrac{15}{8} $
(8.) GCSE Based on Question $6$
A different sequence follows the same rule.
The $1st$ term is $4$
The $3rd$ term is $9.5$
$4\:\:\:\:\:\:\:.......\:\:\:\:\:\:\:9.5$
Work out the $2nd$ term.


$ 4,\:\:\:\:\:\:\:2nd,\:\:\:\:\:\:\:9.5 \\[3ex] 9.5 = \dfrac{4 + 2nd}{2} \\[5ex] 2(9.5) = 4 + 2nd \\[3ex] 19 = 4 + 2nd \\[3ex] 4 + 2nd = 19 \\[3ex] 2nd = 19 - 4 \\[3ex] 2nd\:\:term = 15 $
(9.) JAMB The $nth$ term of the progression

$\dfrac{4}{2},\:\:\dfrac{7}{3},\:\:\dfrac{10}{4},\:\:\dfrac{13}{5}...$ is

$ A.\:\: \dfrac{3n + 1}{n + 1} \\[5ex] B.\:\: \dfrac{3n + 1}{n - 1} \\[5ex] C.\:\: \dfrac{3n - 1}{n + 1} \\[5ex] D.\:\: \dfrac{1 - 3n}{n + 1} \\[5ex] $

$ 1st\:\:term = a = \dfrac{4}{2} = 2 \\[5ex] Test\:\:each\:\:option \\[3ex] Option\:\:A \\[3ex] nth\:\:term = \dfrac{3n + 1}{n + 1} \\[5ex] For\:\: n = 1, \implies a = 2 \\[3ex] nth\:\:term = \dfrac{3(1) + 1}{1 + 1} \\[5ex] = \dfrac{3 + 1}{2} \\[5ex] = \dfrac{4}{2} \\[5ex] = 2 \checkmark \\[3ex] For\:\: n = 2, \implies second\:\:term = \dfrac{7}{3} \\[5ex] nth\:\:term = \dfrac{3(2) + 1}{2 + 1} \\[5ex] = \dfrac{6 + 1}{3} \\[5ex] = \dfrac{7}{3} \checkmark \\[5ex] For\:\: n = 3, \implies third\:\:term = \dfrac{10}{4} \\[5ex] nth\:\:term = \dfrac{3(3) + 1}{3 + 1} \\[5ex] = \dfrac{9 + 1}{4} \\[5ex] = \dfrac{10}{4} \checkmark \\[5ex] For\:\: n = 4, \implies fourth\:\:term = \dfrac{13}{5} \\[5ex] nth\:\:term = \dfrac{3(4) + 1}{4 + 1} \\[5ex] = \dfrac{12 + 1}{5} \\[5ex] = \dfrac{13}{5} \checkmark \\[5ex] Option\:\:A\:\:is\:\:the\:\:correct\:\:option $
(10.) ACT The recursive formula for a sequence is given below, where $a_n$ is the value of the $nth$ term.

$ a_1 = 10 \\[3ex] a_n = a_{n - 1} + 5 \\[3ex] $ Which of the following equations is an explicit formula for this sequence?

$ A.\:\: a_n = -5n + 10 \\[3ex] B.\:\: a_n = 5n + 5 \\[3ex] C.\:\: a_n = 5n + 10 \\[3ex] D.\:\: a_n = 10n - 5 \\[3ex] E.\:\: a_n = 10n + 5 \\[3ex] $

Recursive Seqeunce

$ a_1 = 10 \\[3ex] a_n = a_{n - 1} + 5 \\[3ex] a_2 = a_{2 - 1} + 5 \\[3ex] a_2 = a_1 + 5 = 10 + 5 = 15 \\[3ex] a_3 = a_{3 - 1} + 5 \\[3ex] a_3 = a_2 + 5 = 15 + 5 = 20 \\[3ex] a_1, a_2, a_3 = 10, 15, 20...arithmetic\:\:sequence \\[3ex] a = a_1 = 10 \\[3ex] d = 15 - 10 = 5 \\[3ex] a_n = a + d(n - 1) \\[3ex] a_n = 10 + 5(n - 1) \\[3ex] a_n = 10 + 5n - 5 \\[3ex] a_n = 5n + 5 $
(11.) ACT The first $5$ terms of a sequence are given in the table below.
The sequence is defined by setting $a_1 = 9$ and $a_n = a_{n - 1} + (n - 1)^2$ for $n \ge 2$
What is the sixth term, $a_6$, of this sequence?

$a_1$ $a_2$ $a_3$ $a_4$ $a_5$ $a_6$
$9$ $10$ $14$ $23$ $39$ $?$

$ A.\:\: 62 \\[3ex] B.\:\: 64 \\[3ex] C.\:\: 76 \\[3ex] D.\:\: 78 \\[3ex] E.\:\: 95 \\[3ex] $

Recursive Seqeunce

$ a_n = a_{n - 1} + (n - 1)^2 \\[3ex] a_6 = a_{6 - 1} + (6 - 1)^2 \\[3ex] a_6 = a_{5} + 5^2 \\[3ex] a_6 = 39 + 25 \\[3ex] a_6 = 64 $
(12.) JAMB The $nth$ terms of two sequences are $Q_n = 3 * 2^{n - 2}$ and $U_m = 3 * 2^{2m - 3}$.
Find the product of $Q_2$ and $U_2$

$ A.\:\: 12 \\[3ex] B.\:\: 18 \\[3ex] C.\:\: 6 \\[3ex] D.\:\: 3 \\[3ex] $

$ Q_n = 3 * 2^{n - 2} \\[3ex] Q_2 = 3 * 2^{2 - 2} \\[3ex] Q_2 = 3 * 2^0 \\[3ex] Q_2 = 3 * 1 \\[3ex] Q_2 = 3 \\[3ex] U_m = 3 * 2^{2m - 3} \\[3ex] U_2 = 3 * 2^{2(2) - 3} \\[3ex] U_2 = 3 * 2^{4 - 3} \\[3ex] U_2 = 3 * 2^1 \\[3ex] U_2 = 3 * 2 \\[3ex] U_2 = 6 \\[3ex] Q_2 * U_2 = 3(6) = 18 $
(13.) ACT

Notice the pattern

$ 1 + 3\:\:are\:\:2\:\:numbers \\[3ex] 1 + 3 = 4 = 2^2 \\[3ex] 1 + 3 + 5\:\:are\:\:3\:\:numbers \\[3ex] 1 + 3 + 5 = 9 = 3^3 \\[3ex] 1 + 3 + 5 + 7 \:\:are\:\:4\:\:numbers \\[3ex] 1 + 3 + 5 + 7 = 16 = 4^2 \\[3ex] 1 + 3 + 5 + 7 + 9 \:\:are\:\:5\:\:numbers \\[3ex] 1 + 3 + 5 + 7 + 9 = 5^2 \\[3ex] Similarly \\[3ex] For\:\:144 \\[3ex] 144 = 12^2 \\[3ex] We\:\:need\:\:12\:\:numbers \\[3ex] 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 \:\:are\:\:12\:\:numbers \\[3ex] 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12^2 \\[3ex] Highest\:\:odd\:\:number = 23 $
(14.) ACT What is the $7th$ term in this sequence of "triangular" numbers, defined by the figures below: $1, 3, 6, 10,...?$
Question 14 - OS

$ F.\:\: 7 \\[3ex] G.\:\: 22 \\[3ex] H.\:\: 25 \\[3ex] J.\:\: 28 \\[3ex] K.\:\: 40 \\[3ex] $

Triangular Seqeunce

$ TS_n = \dfrac{n(n + 1)}{2} \\[5ex] TS_7 = \dfrac{7(7 + 1)}{2} \\[5ex] TS_7 = \dfrac{7 * 8}{2} \\[5ex] TS_7 = 7(4) \\[3ex] TS_7 = 28 $