Number patterns & sequences — Mathematics Grade 12

6 exam-style question sets on Number patterns & sequences (Paper 1), each with a hint and a fully worked answer. The app holds 30 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Arithmetic series & quadratic patternsGeometric series, sigma notation & sum to infinityGeometric sequences & seriesGeometric series & sigma notationQuadratic patterns

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Question 1

Arithmetic series & quadratic patternsRoutine16 marksPaper 1

Consider the arithmetic series $11 + 19 + 27 + 35 + \dots$

A quadratic number pattern $T_{1}; T_{2}; T_{3}; \dots$ has first term $T_{1} = 4$ . The second term is 5 more than the first term, and the third term is 9 more than the second term, so $T_{2} - T_{1} = 5$ and $T_{3} - T_{2} = 9$ .

aDetermine the 45th term of the arithmetic series. (3)
bCalculate the sum of the first 45 terms of the series. (2)
cWhich term of the series is equal to 603? (3)
dShow that $T_{6} = 69$ . (2)
eHence prove that the $n$ th term of this pattern is given by $T_{n} = 2 n^{2} - n + 3$ . (3)
fProve that the quadratic pattern is increasing for ALL natural numbers $n$ , that is, that $T_{n + 1} > T_{n}$ for every $n \in N$ . (3)

Hint

For the series use $T_{n} = a + (n - 1) d$ and $S_{n} = \frac{n}{2} (a + T_{n})$ . For the pattern, the first differences $5; 9; \dots$ form an arithmetic sequence whose common difference is the second difference, which equals $2 a$ . For the proof, simplify $T_{n + 1} - T_{n}$ and explain why the result is positive whenever $n \geq 1$ .

Worked answer

a) $a = 11$ and $d = 19 - 11 = 8$ ✓
$T_{45} = 11 + (45 - 1) (8)$ ✓ $= 11 + 352 = 363$ ✓

b) $S_{45} = \frac{45}{2} (11 + 363)$ ✓ $= \frac{45}{2} (374) = 8415$ ✓

c) $11 + (n - 1) (8) = 603$ ✓ $\Rightarrow 8 (n - 1) = 592$ ✓ $\Rightarrow n - 1 = 74 \Rightarrow n = 75$ ✓
The 75th term is 603.

d) First differences: $5; 9; 13; 17; 21$ (second difference 4, so each first difference grows by 4) ✓
$T_{1} = 4; T_{2} = 9; T_{3} = 18; T_{4} = 31; T_{5} = 48; T_{6} = 48 + 21 = 69$ ✓

e) Second difference $= 9 - 5 = 4$ , so $2 a = 4 \Rightarrow a = 2$ ✓
$3 a + b = 5$ (the first of the first differences) $\Rightarrow 6 + b = 5 \Rightarrow b = - 1$ ✓
$a + b + c = T_{1} = 4 \Rightarrow 2 - 1 + c = 4 \Rightarrow c = 3$ ✓
$∴ T_{n} = 2 n^{2} - n + 3$ (check: $T_{6} = 2 (36) - 6 + 3 = 69$ )

f) $T_{n + 1} - T_{n} = [2 (n + 1)^{2} - (n + 1) + 3] - [2 n^{2} - n + 3]$ ✓
$= 2 n^{2} + 4 n + 2 - n - 1 + 3 - 2 n^{2} + n - 3 = 4 n + 1$ ✓
For every natural number $n \geq 1$ we have $4 n + 1 \geq 5 > 0$ , so $T_{n + 1} > T_{n}$ for ALL $n \in N$ . The pattern is therefore increasing ✓
(OR: the first differences $5; 9; 13; \dots$ form an arithmetic sequence with positive first term and positive common difference, so every first difference is positive.)

Modelled on Nov 2023 Paper 1, Question 2 — same skills, new scenario.

Question 2

Geometric series, sigma notation & sum to infinityProblem-solving10 marksPaper 1

The Grade 12 farewell committee at a school in Polokwane sells raffle tickets. On the first day of the sale 5 tickets are sold, and on each day after that TWICE as many tickets are sold as on the previous day, so the daily numbers of tickets sold form the geometric sequence $5; 10; 20; \dots$

Two other sequences both start with the number $x$ : one is arithmetic, with a common difference of 4, and the other is geometric, with a common ratio of $\frac{3}{4}$ .

aWrite down an expression for $T_{n}$ , the number of tickets sold on day $n$ . (1)
bThe total number of tickets sold in the first $p$ days is given by $\sum_{k = 1}^{p} 5 \cdot 2^{k - 1} = 635$ . Calculate the value of $p$ . (4)
cWhen the sum to infinity of the geometric sequence is subtracted from the sum of the first eight terms of the arithmetic sequence, the result is 140. Calculate the value of $x$ . (5)

Hint

A geometric sequence has $T_{n} = a r^{n - 1}$ . In the sigma part, use $S_{p} = \frac{a ( r ^{p} - 1 )}{r - 1}$ and write both sides as powers of 2. In the last part, write $S_{8}$ of the arithmetic sequence and $S_{\infty} = \frac{a}{1 - r}$ of the geometric sequence, both in terms of $x$ , then form an equation.

Worked answer

a) $a = 5$ and $r = 2$ , so $T_{n} = 5 \cdot 2^{n - 1}$ ✓

b) $\sum_{k = 1}^{p} 5 \cdot 2^{k - 1}$ is the sum of the first $p$ terms of the geometric series with $a = 5$ and $r = 2$ :
$S_{p} = \frac{5 ( 2 ^{p} - 1 )}{2 - 1} = 5 (2^{p} - 1)$ ✓
$5 (2^{p} - 1) = 635$ ✓ $\Rightarrow 2^{p} - 1 = 127 \Rightarrow 2^{p} = 128$ ✓
$2^{p} = 2^{7} \Rightarrow p = 7$ ✓ (Check: $5 + 10 + 20 + 40 + 80 + 160 + 320 = 635$ .)

c) Arithmetic sequence: $S_{8} = \frac{8}{2} [2 x + 7 (4)] = 4 (2 x + 28) = 8 x + 112$ ✓
Geometric sequence: $- 1 < r = \frac{3}{4} < 1$ , so $S_{\infty} = \frac{x}{1 - \frac{3}{4}} = 4 x$ ✓
$S_{8} - S_{\infty} = 140$ : $8 x + 112 - 4 x = 140$ ✓
$4 x = 28$ ✓ $\Rightarrow x = 7$ ✓
(Check: $S_{8} = 168$ and $S_{\infty} = 28$ ; $168 - 28 = 140$ .)

Modelled on Nov 2023 Paper 1, Question 3 — same skills, new scenario.

Question 3

Arithmetic series & quadratic patternsComplex14 marksPaper 1

The arithmetic series $5 + 9 + 13 + \dots$ has 20 terms.

A quadratic number pattern $T_{1}; T_{2}; T_{3}; \dots$ is also given. Its first differences form the arithmetic sequence $7; 13; 19; \dots$ and it is further given that $T_{40} = 4725$ .

aCalculate the sum of the 20 terms of the arithmetic series. (2)
bAnother 15 terms are added to the arithmetic series. The sum of the 15 added terms is $1695$ . Write this information as an equation in sigma notation. (4)
cDetermine the value of $T_{39}$ . (3)
dDetermine $T_{n}$ , the general term of the quadratic pattern, in the form $T_{n} = a n^{2} + bn + c$ . (5)

Hint

Use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ for the series; for the pattern, $T_{40} - T_{39}$ equals the 39th term of the first-difference sequence, and the second difference gives $2 a$ .

Worked answer

a) $S_{20} = \frac{20}{2} [2 (5) + 19 (4)]$ ✓ $= 10 (86) = 860$ ✓

b) General term of the series: $T_{k} = 5 + (k - 1) (4) = 4 k + 1$ ✓✓ The added terms are terms 21 to 35 ✓, so $\sum_{k = 21}^{35} (4 k + 1) = 1695$ ✓

c) The first differences form an arithmetic sequence with first term 7 and common difference 6, so the difference between $T_{39}$ and $T_{40}$ is the 39th first difference: $7 + 38 (6) = 235$ ✓✓
$T_{39} = 4725 - 235 = 4490$ ✓

d) Second difference $= 6$ , so $2 a = 6 \Rightarrow a = 3$ ✓
$3 a + b = 7$ (first term of the first differences) $\Rightarrow b = - 2$ ✓
$T_{40} = 3 (40)^{2} - 2 (40) + c = 4725$ ✓ $\Rightarrow 4720 + c = 4725 \Rightarrow c = 5$ ✓
$T_{n} = 3 n^{2} - 2 n + 5$ ✓

Modelled on Nov 2024 Paper 1, Question 2 — same skills, new scenario.

Question 4

Geometric sequences & seriesComplex10 marksPaper 1

A glazier cuts a sequence of square glass tiles for a mosaic. The first tile has sides of length 16 cm, and the sides of each tile after the first are HALF the length of the sides of the previous tile, so the side lengths (in cm) form the geometric sequence $16; 8; 4; \dots$ The sequence of tiles continues in this way indefinitely.

aWrite down the side length of the FIFTH tile. (2)
bCalculate the total area of the first 10 tiles. (4)
cOne of the tiles has sides of length $\frac{1}{64}$ cm. Which tile is it? Show ALL working. (4)

Hint

The areas of the tiles form a geometric series whose common ratio is the SQUARE of the side-length ratio; for the last part, write both sides of the equation as powers of 2.

Worked answer

a) $r = \frac{1}{2}$ , so the fifth side length is $16 (\frac{1}{2})^{4}$ ✓ $= 1$ cm ✓

b) Areas (in cm²): $256; 64; 16; \dots$ — a geometric series with $a = 256$ and $r = (\frac{1}{2})^{2} = \frac{1}{4}$ ✓✓
$S_{10} = \frac{256 [ 1 - ( \frac{1}{4} ) ^{10} ]}{1 - \frac{1}{4}}$ ✓ $= 341, 33$ cm² ✓

c) $16 (\frac{1}{2})^{n - 1} = \frac{1}{64}$ ✓
$2^{4} \cdot 2^{- (n - 1)} = 2^{- 6}$ ✓ $\Rightarrow 2^{5 - n} = 2^{- 6}$ ✓
$5 - n = - 6 \Rightarrow n = 11$ ✓ ( $n \in N$ ) — it is the 11th tile.

Modelled on Nov 2024 Paper 1, Question 3 — same skills, new scenario.

Question 5

Geometric series & sigma notationComplex15 marksPaper 1

An infinite geometric series has, as its first three terms, $(t + 25)$ ; $(t + 9)$ and $(t + 1)$ , where $t$ is a constant.

The arithmetic sequence $5; 7; 9; \dots$ has general term $T_{p} = 2 p + 3$ . It is further given that $\sum_{p = k}^{131} (2 p + 3) = 17248$ , where $k$ is a natural number.

aShow that $t = 7$ . (3)
bDetermine the 25th term of the series. Write the answer in the form $2^{x}$ . (3)
cCalculate the sum to infinity, $S_{\infty}$ , of this series. (2)
dWrite down the value of $T_{100} - T_{61}$ . (2)
eCalculate the value of $k$ . (5)

Hint

Equate the ratios $\frac{T _{2}}{T _{1}}$ and $\frac{T _{3}}{T _{2}}$ to find $t$ ; for the last part, the sum from $p = k$ to 131 has $132 - k$ terms.

Worked answer

a) Constant ratio: $\frac{t + 9}{t + 25} = \frac{t + 1}{t + 9}$ ✓ $\Rightarrow (t + 9)^{2} = (t + 25) (t + 1)$
$t^{2} + 18 t + 81 = t^{2} + 26 t + 25$ ✓ $\Rightarrow 8 t = 56 \Rightarrow t = 7$ ✓

b) With $t = 7$ the terms are $32; 16; 8$ , so $r = \frac{1}{2}$ ✓
$T_{25} = 32 (\frac{1}{2})^{24}$ ✓ $= 2^{5} \cdot 2^{- 24} = 2^{- 19}$ ✓

c) Since $- 1 < r < 1$ : $S_{\infty} = \frac{32}{1 - \frac{1}{2}}$ ✓ $= 64$ ✓

d) $T_{100} - T_{61} = [2 (100) + 3] - [2 (61) + 3] = 203 - 125$ ✓ $= 78$ ✓ (39 gaps of the common difference 2)

e) Number of terms: $n = 131 - k + 1 = 132 - k$ ✓; first term $2 k + 3$ , last term $2 (131) + 3 = 265$ ✓
$\frac{132 - k}{2} [(2 k + 3) + 265] = 17248 \Rightarrow (132 - k) (k + 134) = 17248$ ✓
$k^{2} + 2 k - 440 = 0$ ✓ $\Rightarrow (k - 20) (k + 22) = 0 \Rightarrow k = 20$ ✓ (reject $k = - 22$ , since $k \in N$ )

Modelled on Nov 2025 Paper 1, Question 2 — same skills, new scenario.

Question 6

Quadratic patternsComplex10 marksPaper 1

A model rocket is fired upwards from a launch platform. An on-board altimeter logs the height of the rocket above the ground, in metres, every second after lift-off. The logged heights form a quadratic number pattern, where $T_{n}$ is the height, in metres, $n$ seconds after lift-off. The first three readings are:

Time $n$ (seconds)	1	2	3
Height $T_{n}$ (metres)	36	58	76

aWrite down the heights logged at $n = 4$ and at $n = 5$ . (2)
bShow that $T_{n} = - 2 n^{2} + 28 n + 10$ . (3)
cCalculate the maximum height, in metres, that the rocket reaches according to this pattern. (3)
dThe rocket is at a height of 90 m at TWO different times. Calculate the value of $n$ at the SECOND of these times. (2)

Hint

Use the constant second difference to extend the readings; the vertex $n = - \frac{b}{2 a}$ gives the maximum, and the symmetry of the parabola locates the second time.

Worked answer

a) First differences $22; 18$ , so the second difference is $- 4$ and the next differences are $14$ and $10$ .
$T_{4} = 76 + 14 = 90$ m ✓ and $T_{5} = 90 + 10 = 100$ m ✓

b) Second difference $= - 4$ , so $2 a = - 4 \Rightarrow a = - 2$ ✓
$3 a + b = 22 \Rightarrow b = 28$ ✓
$a + b + c = 36 \Rightarrow - 2 + 28 + c = 36 \Rightarrow c = 10$ ✓
$T_{n} = - 2 n^{2} + 28 n + 10$ (check: $T_{3} = - 18 + 84 + 10 = 76$ )

c) Maximum at $n = - \frac{b}{2 a} = - \frac{28}{2 ( - 2 )} = 7$ ✓✓
$T_{7} = - 2 (7)^{2} + 28 (7) + 10 = 108$ m ✓

d) $- 2 n^{2} + 28 n + 10 = 90 \Rightarrow n^{2} - 14 n + 40 = 0 \Rightarrow (n - 4) (n - 10) = 0$ ✓
$n = 4$ or $n = 10$ ; the SECOND time is at $n = 10$ seconds ✓

Modelled on Nov 2025 Paper 1, Question 3 — same skills, new scenario.

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