Solved Examples on Other Sequences

Formulas: Formulas

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 a wrong answer.

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.
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.) JAMB Find the $nth$ term of the sequence

$ \;\;\;\;\;\;\;\;\;\;3, 6, 10, 15, 21,... \\[3ex] A.\;\; n\left(n - \dfrac{1}{2}\right) \\[5ex] B.\;\; n\left(n + \dfrac{1}{2}\right) \\[5ex] C.\;\; \dfrac{(n + 1)(n + 2)}{2} \\[5ex] D.\;\; n(2n + 1) \\[3ex] $

The fastest way to solve this question (due to the JAMB timed exam) is to test each option
Let us test for the first term (when $n$ = 1)

$ a = 3 \\[3ex] \underline{Test\;\;each\;\;option\;\;for\;\;a} \\[3ex] When\;\;n = 1, \;\;\; a = 3 \\[3ex] (A.)\;\; n\left(n - \dfrac{1}{2}\right) \\[5ex] 1\left(1 - \dfrac{1}{2}\right) \\[5ex] 1 * \dfrac{1}{2} = \dfrac{1}{2} \unicode{x2718} \\[3ex] \dfrac{1}{2} \ne 3...eliminate \\[5ex] (B.)\;\; n\left(n + \dfrac{1}{2}\right) \\[5ex] n\left(n + \dfrac{1}{2}\right) \\[5ex] 1\left(1 + \dfrac{1}{2}\right) \\[5ex] 1 * \left(\dfrac{2}{2} + \dfrac{1}{2}\right) \\[5ex] 1 * \dfrac{3}{2} = \dfrac{3}{2} \unicode{x2718} \\[3ex] \dfrac{3}{2} \ne 3...eliminate \\[5ex] (C.)\;\; \dfrac{(n + 1)(n + 2)}{2} \\[5ex] \dfrac{(1 + 1)(1 + 2)}{2} \\[5ex] \dfrac{2(3)}{2} = 3 \checkmark \\[5ex] (D.)\;\; n(2n + 1) \\[3ex] 1(2(1) + 1) \\[3ex] 1(2 + 1) \\[3ex] 1(3) = 3 \checkmark \\[3ex] Options\;\;C\;\;and\;\;D\;\;remain \\[3ex] \underline{Test\;\;for\;\;n = 2} \\[3ex] When\;\;n = 2, \;\;\; 2nd\;\;term = 6 \\[3ex] (C.)\;\; \dfrac{(n + 1)(n + 2)}{2} \\[5ex] \dfrac{(2 + 1)(2 + 2)}{2} \\[5ex] \dfrac{(3)(4)}{2} = 3(2) = 6 \checkmark \\[5ex] (D.)\;\; n(2n + 1) \\[3ex] 2(2(2) + 1) \\[3ex] 2(4 + 1) \\[3ex] 2(5) = 10 \unicode{x2718} \\[3ex] 10 \ne 6 ...eliminate \\[3ex] Correct\;\;Option\;\;is\;\;C \\[3ex] nth\;\;term = \dfrac{(n + 1)(n + 2)}{2}...Quadratic\;\;Sequence $

(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.) ACT There is a pattern when adding the cubes of the first c consecutive counting numbers, as illustrated below.

$ ~~~~~~~~~~ 1^3 + 2^3 = 9 = (1 + 2)^2 \\[3ex] ~~~~~~~~~~ 1^3 + 2^3 + 3^3 = 36 = (1 + 2 + 3)^2 \\[3ex] $ Which of the following is an expression for the sum of the cubes of the first c consecutive counting numbers?

$ F.\;\; (c + 1)^3 \\[3ex] G.\;\; (c + 1)^2 \\[3ex] H.\;\; (1 + 2 + ... + c)^c \\[3ex] J.\;\; (1 + 2 + ... + c)^3 \\[3ex] K.\;\; (1 + 2 + ... + c)^2 \\[3ex] $

$ ~~~~~~~~~~ 1^3 + 2^3 = 9 = (1 + 2)^2 \\[3ex] ~~~~~~~~~~ 1^3 + 2^3 + 3^3 = 36 = (1 + 2 + 3)^2 \\[3ex] Similarly: \\[3ex] ~~~~~~~~~~ 1^3 + 2^3 + ... + c^3 = (1 + 2 + ... + c)^2 $ The correct answer is Option K.

(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.)

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.) JAMB What is the $nth$ term of the sequence $2, 6, 12, 20, ...?$

$ A.\;\; 4n - 2 \\[3ex] B.\;\; 2(3n - 1) \\[3ex] C.\;\; n^2 + n \\[3ex] D.\;\; n^2 + 3n + 2 \\[3ex] $

The fastest way to solve this question (due to the JAMB timed exam) is to test each option
Let us test for the first term (when $n$ = 1)

$ a = 2 \\[3ex] \underline{Test\;\;each\;\;option\;\;for\;\;a} \\[3ex] When\;\;n = 1, \;\;\; a = 2 \\[3ex] (A.)\;\; 4n - 2 \\[3ex] 4(1) - 2 \\[3ex] 4 - 2 = 2 \checkmark \\[3ex] (B.)\;\; 2(3n - 1) \\[3ex] 2(3(1) - 1) \\[3ex] 2(3 - 1) \\[3ex] 2(2) = 4 \unicode{x2718} \\[3ex] 4 \ne 2...eliminate \\[3ex] (C.)\;\; n^2 + n \\[3ex] 1^2 + 1 \\[3ex] 1 + 1 = 2 \checkmark \\[3ex] (D.)\;\; n^2 + 3n + 2 \\[3ex] 1^2 + 3(1) + 2 \\[3ex] 1 + 3 + 2 = 6 \unicode{x2718} \\[3ex] 6 \ne 2...eliminate \\[3ex] Options\;\;A\;\;and\;\;C\;\;remain \\[3ex] \underline{Test\;\;for\;\;n = 2} \\[3ex] When\;\;n = 2, \;\;\; 2nd\;\;term = 6 \\[3ex] (A.)\;\; 4n - 2 \\[3ex] 4(2) - 2 \\[3ex] 8 - 2 = 6 \checkmark \\[3ex] (C.)\;\; n^2 + n \\[3ex] 2^2 + 2 \\[3ex] 4 + 2 = 6 \checkmark \\[3ex] \underline{Test\;\;for\;\;n = 3} \\[3ex] When\;\;n = 3, \;\;\; 3rd\;\;term = 12 \\[3ex] (A.)\;\; 4n - 2 \\[3ex] 4(3) - 2 \\[3ex] 12 - 2 = 10 \unicode{x2718} \\[3ex] 10 \ne 12...eliminate \\[3ex] (C.)\;\; n^2 + n \\[3ex] 3^2 + 3 \\[3ex] 9 + 3 = 12 \checkmark \\[3ex] Correct\;\;Option\;\;is\;\;C \\[3ex] nth\;\;term = n^2 + n...Quadratic\;\;Sequence $

(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 The first $4$ elements of a pattern are shown below.
Each element is composed of small squares that are $18$ inches wide and $18$ inches long.
Each element is a square with both dimensions $18$ inches less than the dimensions of the next element.
What is the perimeter, in feet, of the 5th element?
Question 13

$ F.\;\; 6 \\[3ex] G.\;\; 7.5 \\[3ex] H.\;\; 20 \\[3ex] J.\;\; 25 \\[3ex] K.\;\; 30 \\[3ex] $

Notice the pattern:

$ 1\;\;square = 1^2 \\[3ex] 4\;\; squares = 2^2 \\[3ex] 9\;\; squares = 3^2 \\[3ex] 16\;\; squares = 4^2 \\[3ex] $ This is a Square sequence
Next pattern would be $25$ squares because of $5^2$
Side of each square = $18$ inches
Length of next pattern = 5 square sides (outer) = (5 * 18) inches
Width of next pattern = 5 squares sides (outer) = (5 * 18) inches

$ Perimeter\;\;of\;\;next\;\;pattern \\[3ex] = (2 * Length) + (2 * Width) \\[3ex] = (2 * 5 * 18) + (2 * 5 * 18) \\[3ex] = 180 + 180 \\[3ex] = 360\;inches \\[3ex] 12\;inches = 1\;\;foot \\[3ex] \therefore 360\;inches \\[3ex] = \dfrac{360}{12} \;feet \\[5ex] = 30\;feet $

(16.) JAMB The $nth$ term of a sequence is given by $3^{1 - n}$
Find the sum of the first three terms of the sequence.

$ A.\:\: \dfrac{13}{9} \\[5ex] B.\:\: 1 \\[3ex] C.\:\: \dfrac{1}{3} \\[5ex] D.\:\: \dfrac{1}{9} \\[5ex] $

$ nth\;\;term = 3^{1 - n} \\[3ex] 1st\;\;term \implies n = 1 \\[3ex] = 3^{1 - 1} \\[3ex] = 3^0 \\[3ex] = 1 \\[3ex] 2nd\;\;term \implies n = 2 \\[3ex] = 3^{1 - 2} \\[3ex] = 3^{-1} \\[3ex] = \dfrac{1}{3^1} \\[5ex] = \dfrac{1}{3} \\[5ex] 3rd\;\;term \implies n = 3 \\[3ex] = 3^{1 - 3} \\[3ex] = 3^{-2} \\[3ex] = \dfrac{1}{3^2} \\[5ex] = \dfrac{1}{9} \\[5ex] Sum\;\;of\;\;the\;\;first\;\;terms \\[3ex] = 1 + \dfrac{1}{3} + \dfrac{1}{9} \\[5ex] = \dfrac{9}{9} + \dfrac{3}{9} + \dfrac{1}{9} \\[5ex] = \dfrac{13}{9} $