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These are the solutions to Cambridge Ordinary Level (O-Level) past questions on Sequences.
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$ (1.)\:\: AS_n = a + d(n - 1) \\[5ex] (2.)\:\: AS_n = vn + w \:\:where\:\: v = d \:\:and\:\: w = a - d \\[5ex] (3.)\:\: p = a + d(n - 1) \\[5ex] (4.)\:\: SAS_n = \dfrac{n}{2}(a + AS_n) \\[7ex] (5.)\:\: SAS_n = \dfrac{n}{2}(a + p) \\[7ex] (6.)\:\: SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[7ex] (7.)\:\: n = \dfrac{2 * SAS_n}{a + p} \\[7ex] (8.)\:\: n = \dfrac{p - a + d}{d} \\[7ex] (9.)\:\: n = \dfrac{-(2a - d) \pm \sqrt{(2a - d)^2 + 8d*SAS_n}}{2d} \\[7ex] (10.)\;\; d = \dfrac{(p - a)(p + a)}{2 * SAS_n - p - a} $


$ (1.)\:\: GS_n = ar^{n - 1} \\[5ex] (2.)\:\: SGS_n = \dfrac{a(r^{n} - 1)}{r - 1} \:\:for\:\: r \gt 1 \\[7ex] (3.)\:\: SGS_n = \dfrac{a(1 - r^{n})}{1 - r} \:\:for\:\: r \lt 1 \\[7ex] (4.)\:\: n = \dfrac{\log{\left[\dfrac{SGS_n(r - 1)}{a} + 1\right]}}{\log r} \\[7ex] (5.)\:\: If\:\:r \lt 1,\:\:the\:\:series\:\:converges\:\:and\:\: S_{\infty} = \dfrac{a}{1 - r} \\[7ex] (6.)\:\: If\:\:r \gt 1,\:\:the\:\:series\:\:diverges \\[5ex] (7.)\:\: If\:\:r = 1,\:\:S_{\infty}\:\:DNE \\[5ex] (8.)\:\: r = \dfrac{S_{\infty} - a}{S_{\infty}} \\[7ex] (9.)\:\: a = S_{\infty}(1 - r) $


$ QS = 1st,\:\:\:\:2nd,\:\:\:3rd,\:\:\:4th,... \\[5ex] QS_n = an^2 + bn + c \\[5ex] (1.)\:\: a = \dfrac{1st + 3rd - 2(2nd)}{2} \\[7ex] (2.)\:\: b = \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} \\[7ex] (3.)\:\: c = 3(1st) - 3(2nd) + 3rd \\[5ex] (4.)\:\: \therefore QS_n = \dfrac{1st + 3rd - 2(2nd)}{2} * n^2 + \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} * n + 3(1st) - 3(2nd) + 3rd \\[7ex] The\:\:Left\:\:Hand\:\:Side\:\:must\;\;be\:\:equal\:\:to\;\;the\:\:Right\:\:Hand\:\:Side \\[5ex] (5.)\:\: a + b + c = 1st \\[5ex] (6.)\:\: 4a + 2b + c = 2nd \\[5ex] (7.)\:\: 9a + 3b + c = 3rd \\[5ex] (8.)\:\: 3a + b = 2nd - 1st \\[5ex] (9.)\:\: 8a + 2b = 3rd - 1st $


$ \underline{Triangular\;\;Number\;\;Sequence} \\[3ex] (1.)\:\: TS_n = \dfrac{n(n + 1)}{2} \\[7ex] (2.)\;\; n = \dfrac{\sqrt{8 * TS_n + 1} - 1}{2} \\[7ex] (3.)\:\: TS_n = C(n + 1, 2)...Combinatorics\;\;symbol\;\;for\;\;Combinations \\[5ex] (4.)\;\; STS_n = \dfrac{n(n + 1)(n + 2)}{6} \\[7ex] (5.)\:\: STS_n = C(n + 2, 3)...Combinatorics\;\;symbol\;\;for\;\;Combinations \\[7ex] \underline{Square\;\;Number\;\;Sequence} \\[3ex] (1.)\:\: SS_n = n^2 \\[5ex] (2.)\;\; n = \sqrt{SS_n} \\[5ex] (3.)\;\; SSS_n = \dfrac{n(n + 1)(2n + 1)}{6} \\[7ex] \underline{Cube\;\;Number\;\;Sequence} \\[3ex] (1.)\:\: CS_n = n^3 \\[5ex] (2.)\;\; n = \sqrt[3]{CS_n} \\[5ex] (3.)\;\; SCS_n = \left[\dfrac{n(n + 1)}{2}\right]^2 \\[7ex] (4.)\;\; n = \dfrac{\sqrt{8\sqrt{SCS_n} + 1} - 1}{2} \\[7ex] $


$ \underline{First-Order\;\;Linear\;\;Recurrence\;\;Relation} \\[3ex] (1.)\:\: RS_{n + 1} = r * RS_{n} + a \\[5ex] (2.)\:\: RS_{n + 1} = \dfrac{RS_1 * r^n(r - 1) + a(r^n - 1)}{r - 1} \;\;\;for\;\;r \gt 1 \\[7ex] (3.)\;\; RS_{n + 1} = \dfrac{RS_1 * r^n(1 - r) + a(1 - r^n)}{1 - r} \;\;\;for\;\;r \lt 1 \\[7ex] \underline{Fibonacci\;\;Sequence} \\[3ex] (1.)\;\; \phi = \dfrac{1 + \sqrt{5}}{2} \\[7ex] (2.)\;\; FS_n = \dfrac{\sqrt{5}}{5}\left[\left(\dfrac{1 + \sqrt{5}}{2}\right)^n - \left(\dfrac{1 - \sqrt{5}}{2}\right)^n\right] \\[7ex] (3.)\;\; FS_n = \dfrac{\sqrt{5}}{5}\left[\phi^n - \left(\dfrac{1 - \sqrt{5}}{2}\right)^n\right] \\[7ex] (4.)\;\; SFS_n = \dfrac{\sqrt{5}}{5}\left[\dfrac{\phi^3(\phi^{n - 1} - 1) + [(\phi - 1)(1 - \phi^2)][(1 - \phi)^{n - 1} - 1]}{\phi(\phi - 1)}\right] + 1 $

(1.) (a.) The diagrams shows the first four patterns in a sequence.

Number 1-1st

(i.) Draw Pattern 5 on the grid below.

Number 1-2nd

(ii.) Complete the table.
Pattern (n) 1 2 3 4 5 6
Total number of triangles 1 4 9 16
Number of grey triangles 0 1 3
Number of white triangles 1 3 6

(iii.) Write an expression, in terms of n, for the total number of triangles in Pattern n.
(iv.) Write an expression, in terms of n, for the number of white triangles in Pattern n.

(b.) The 3rd term of a linear sequence is 34.
The 8th term of the same linear sequence is 14.
(i.) Find the value of the first term of this sequence.
(ii.) Find the value of the first negative term of this sequence.


(i.) Pattern 5 is drawn as shown:
Number 1

(ii.), (iii.), and (iv.)
Let us derive a formula/rule for the:
Total number of triangles
Number of grey triangles
Number of white triangles

The goal is to study the sequence of each and compare it with the sequence of the pattern.
Then, find the formula for the sequence.
Notice the colors.

$ \text{Pattern Sequence: } \color{darkblue}{1, 2, 3, 4, 5, 6, ..., n} \\[5ex] \text{Total number of triangles: } \color{black}{1, 4, 9, 16, ..., ..., ..., ...} \\[3ex] \color{darkblue}{1}^2 = \color{black}{1} \\[3ex] \color{darkblue}{2}^2 = \color{black}{4} \\[3ex] \color{darkblue}{3}^2 = \color{black}{9} \\[3ex] \color{darkblue}{4}^2 = \color{black}{16} \\[3ex] \color{darkblue}{5}^2 = \color{black}{25} \\[3ex] \color{darkblue}{6}^2 = \color{black}{36} \\[3ex] \color{darkblue}{n}^2 = \color{black}{n^2} \\[3ex] $ This one is pretty straightforward.
Let us look at the next one.
For the number of grey triangles, we do not need the formula for the nthe term
So, let's just look at the pattern and complete the table.

$ \text{Number of grey triangles: } \color{purple}{0, 1, 3, ..., ..., ...} \\[3ex] \color{purple}{0} \\[3ex] \color{purple}{0} + \color{darkblue}{1} = \color{purple}{1} \\[3ex] \color{purple}{1} + \color{darkblue}{2} = \color{purple}{3} \\[3ex] \color{purple}{3} + \color{darkblue}{3} = \color{purple}{6} \\[3ex] \color{purple}{6} + \color{darkblue}{4} = \color{purple}{10} \\[3ex] \color{purple}{10} + \color{darkblue}{5} = \color{purple}{15} \\[3ex] $ For the number of white triangles, let us complete it.
Then, we can determine the formula for the nthe term of the sequence

$ \text{Number of white triangles: } \color{brown}{1, 3, 6, ..., ..., ...} \\[3ex] \color{brown}{1} \\[3ex] \color{brown}{1} + \color{darkblue}{2} = \color{brown}{3} \\[3ex] \color{brown}{3} + \color{darkblue}{3} = \color{brown}{6} \\[3ex] \color{brown}{6} + \color{darkblue}{4} = \color{brown}{10} \\[3ex] \color{brown}{10} + \color{darkblue}{5} = \color{brown}{15} \\[3ex] \color{brown}{15} + \color{darkblue}{6} = \color{brown}{21} \\[3ex] \color{brown}{1, 3, 6, 10, 15, 21} \text{ is a triangular sequence} \\[3ex] \text{So, we use the formula for the nth term of a Triangular Sequence} \\[3ex] \text{Number of white triangles} = \dfrac{n(n + 1)}{2} \\[5ex] $ The completed table is:
Pattern (n) 1 2 3 4 5 6
Total number of triangles 1 4 9 16 25 36
Number of grey triangles 0 1 3 6 10 15
Number of white triangles 1 3 6 10 15 21

Student: Mr. C
I have a question.
Teacher: Okay..., ask it
Student: Do I need to memorize the formula for the nth term of a triangular sequence?
Teacher: Not necessarily
Student: How did you know that the number of white triangles is a triangular sequence?
Teacher: I knew it because of the pattern.
Student: So, I guess you memorized it?
Teacher: Well, I have solved a lot of questions on Sequences...so I am familiar with it.
Student: Assume I do not know that it is a triangular sequence.
How would I answer that question?
Teacher: First, we check whether it is a linear sequence.
Student: It is not a linear sequence: $6 - 3 \ne 3 - 1$
Teacher" Next, we check whether it is a quadratic sequence
Student: Okay.

$ 1, 3, 6, 10, 15, 21 \\[3ex] \underline{\text{1st Difference}} \\[3ex] 3 - 1 = 2 \\[3ex] 6 - 3 = 3 \\[3ex] 10 - 6 = 4 \\[3ex] 15 - 10 = 5 \\[5ex] \underline{\text{2nd Difference}} \\[3ex] 3 - 2 = 1 \\[3ex] 4 - 3 = 1 \\[3ex] 5 - 4 = 1 \\[3ex] $ The 2nd differences are the same.
It is a Quadratic Sequence
Teacher: Okay, let's take it from here.

$ 1, 3, 6, 10, 15, 21 \\[3ex] 1st, 2nd, 3rd, ... \\[3ex] \text{Memorize these equations. See the pattern for easy memorization} \\[3ex] a + b + c = 1st = 1 ...eqn.(1) \\[3ex] 4a + 2b + c = 2nd = 3 ...eqn.(2) \\[3ex] 9a + 3b + c = 3rd = 6 ...eqn.(3) \\[5ex] eqn.(2) - eqn.(1) \implies \\[3ex] 4a + 2b + c - (a + b + c) = 3 - 1 \\[3ex] 4a + 2b + c - a - b - c = 2 \\[3ex] 3a + b = 2 ...eqn.(4) \\[3ex] eqn.(3) - eqn.(2) \implies \\[3ex] 9a + 3b + c - (4a + 2b + c) = 6 - 3 \\[3ex] 9a + 3b + c - 4a - 2b - c = 3 \\[3ex] 5a + b = 3 ...eqn.(5) \\[3ex] eqn.(5) - eqn.(4) \implies \\[3ex] 5a + b - (3a + b) = 3 - 2 \\[3ex] 5a + b - 3a - b = 1 \\[3ex] 2a = 1 \\[3ex] a = \dfrac{1}{2} \\[5ex] \text{Substitute for a in eqn.(4)} \\[3ex] b = 2 - 3a \\[3ex] b = 2 - 3\left(\dfrac{1}{2}\right) \\[5ex] b = \dfrac{4}{2} - \dfrac{3}{2} \\[5ex] b = \dfrac{1}{2} \\[5ex] \text{Substitute for a and b in eqn.(1)} \\[3ex] c = 1 - a - b \\[3ex] c = 1 - \dfrac{1}{2} - \dfrac{1}{2} \\[5ex] c = 0 \\[3ex] (iv.) \\[3ex] \text{nth term of the sequence} \\[3ex] = an^2 + bn + c \\[3ex] = \dfrac{1}{2}n^2 + \dfrac{1}{2}n \\[5ex] = \dfrac{n^2 + n}{2} \\[5ex] = \dfrac{n(n + 1)}{2} \\[5ex] (b.) \\[3ex] \underline{\text{Linear Sequence}} \\[3ex] a = \text{first term} \\[3ex] d = \text{common difference} \\[3ex] n = \text{number of terms} \\[3ex] (i.) \\[3ex] \text{3rd term} = a + 2d = 34 ...eqn.(1) \\[3ex] \text{8th term} = a + 7d = 14 ...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \implies \\[3ex] a + 7d - (a + 2d) = 14 - 34 \\[3ex] a + 7d - a - 2d = -20 \\[3ex] 5d = -20 \\[3ex] d = -\dfrac{20}{5} \\[5ex] d = -4 \\[3ex] \text{Substitute for d in eqn.(1)} \\[3ex] a = 34 - 2d \\[3ex] a = 34 - 2(-4) \\[3ex] a = 34 + 8 \\[3ex] a = 42 \\[5ex] (ii.) \\[3ex] \text{negative} \implies \lt 0 \\[3ex] a + nd \lt 0 \\[3ex] 42 - 4n \lt 0 \\[3ex] 42 \lt 4n \\[3ex] 4n \gt 42 \\[3ex] n \gt \dfrac{42}{4} \\[5ex] n \gt 10.5 \\[3ex] n = 11...\text{smallest value of n in this context} \\[3ex] \text{12th term: } a + 11d \\[3ex] = 42 + 11(-4) \\[3ex] = 42 - 44 \\[3ex] = -2 ...\text{first negative term of the sequence} $
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