0 / 20 questions attempted

Patterns & Sequences

IEB Grade 12 Mathematics — Paper 1 (~25 marks)

IEB Information Sheet — Your Formulas
Arithmetic: nth term \( T_n = a + (n-1)d \)
Arithmetic: sum of n terms \( S_n = \frac{n}{2}[2a + (n-1)d] \)
Geometric: nth term \( T_n = ar^{n-1} \)
Geometric: sum of n terms \( S_n = \frac{a(r^n - 1)}{r - 1}, \; r \neq 1 \)
Convergent geometric: sum to infinity \( S_\infty = \frac{a}{1 - r}, \; -1 < r < 1 \)
Alternative sum formula \( S_n = \frac{n}{2}(a + l) \) where \( l \) is the last term

1. Arithmetic Sequences

An arithmetic sequence is a list of numbers where you add the same value each time to get from one term to the next. That value is called the common difference, and we call it \( d \).

The Key Idea

If you can get from any term to the next by adding (or subtracting) the same number, it's arithmetic.

For example: \( 3; \; 7; \; 11; \; 15; \; \ldots \) — here \( d = 4 \) because each term is 4 more than the last.

Or: \( 20; \; 14; \; 8; \; 2; \; \ldots \) — here \( d = -6 \) (subtracting 6 each time).

The Formula

To find any term directly (without listing them all), use:

$$ T_n = a + (n-1)d $$

where \( a \) = first term, \( d \) = common difference, \( n \) = position of the term.

Worked Example

Find the 20th term of: \( 5; \; 9; \; 13; \; 17; \; \ldots \)

\( a = 5, \quad d = 9 - 5 = 4, \quad n = 20 \)
\( T_{20} = 5 + (20-1)(4) = 5 + 76 = 81 \)
To find \( d \), take any term and subtract the one before it: \( d = T_2 - T_1 \) (or \( T_3 - T_2 \), etc.).
Q1 Easy
Find the 15th term of the sequence: \( 2; \; 7; \; 12; \; 17; \; \ldots \)
Solution
\( a = 2, \quad d = 7 - 2 = 5, \quad n = 15 \)
\( T_{15} = 2 + (15-1)(5) = 2 + 70 = 72 \)
Q2 Medium
The 3rd term of an arithmetic sequence is 11 and the 7th term is 27. Find \( a \) and \( d \).
Solution
From the formula: \( T_3 = a + 2d = 11 \) ... (1)
\( T_7 = a + 6d = 27 \) ... (2)
Subtract (1) from (2): \( 4d = 16 \), so \( d = 4 \)
Substitute back: \( a + 8 = 11 \), so \( a = 3 \)
Q3 Medium
Which term of the sequence \( 100; \; 93; \; 86; \; 79; \; \ldots \) is equal to 2?
Solution
\( a = 100, \quad d = -7 \)
Set \( T_n = 2 \): \( 100 + (n-1)(-7) = 2 \)
\( 100 - 7n + 7 = 2 \)
\( 107 - 7n = 2 \)
\( 7n = 105 \), so \( n = 15 \)
The 15th term is equal to 2.

2. Arithmetic Series

A series is what you get when you add up the terms of a sequence. An arithmetic series is the sum of terms in an arithmetic sequence.

The Sum Formula

$$ S_n = \frac{n}{2}[2a + (n-1)d] $$

This gives you the sum of the first \( n \) terms. If you know the last term \( l \), you can also use:

$$ S_n = \frac{n}{2}(a + l) $$

Worked Example

Find the sum of the first 30 terms of: \( 4; \; 10; \; 16; \; 22; \; \ldots \)

\( a = 4, \quad d = 6, \quad n = 30 \)
\( S_{30} = \frac{30}{2}[2(4) + (30-1)(6)] \)
\( = 15[8 + 174] = 15 \times 182 = 2730 \)
Finding a Single Term from the Sum

A very useful trick: \( T_n = S_n - S_{n-1} \). This means to find the 10th term, compute \( S_{10} - S_9 \).

Q4 Easy
Find the sum of the first 20 terms of: \( 1; \; 4; \; 7; \; 10; \; \ldots \)
Solution
\( a = 1, \quad d = 3, \quad n = 20 \)
\( S_{20} = \frac{20}{2}[2(1) + 19(3)] = 10[2 + 57] = 10 \times 59 = 590 \)
Q5 Medium
An arithmetic series has \( a = 5 \) and \( d = 3 \). How many terms must be added to give a sum of 440?
Solution
\( S_n = \frac{n}{2}[2(5) + (n-1)(3)] = 440 \)
\( \frac{n}{2}[10 + 3n - 3] = 440 \)
\( \frac{n}{2}[3n + 7] = 440 \)
\( n(3n + 7) = 880 \)
\( 3n^2 + 7n - 880 = 0 \)
Using the quadratic formula: \( n = \frac{-7 \pm \sqrt{49 + 10560}}{6} = \frac{-7 \pm \sqrt{10609}}{6} = \frac{-7 \pm 103}{6} \)
\( n = \frac{96}{6} = 16 \) (reject the negative answer)
16 terms must be added.
Q6 Hard
Given that \( S_n = 3n^2 - 2n \), find the value of \( T_5 \) and show that the sequence is arithmetic.
Solution
\( T_5 = S_5 - S_4 \)
\( S_5 = 3(25) - 10 = 65 \)
\( S_4 = 3(16) - 8 = 40 \)
\( T_5 = 65 - 40 = 25 \)
To show it's arithmetic, find \( T_n \) in general:
\( T_n = S_n - S_{n-1} = [3n^2 - 2n] - [3(n-1)^2 - 2(n-1)] \)
\( = 3n^2 - 2n - 3n^2 + 6n - 3 + 2n - 2 = 6n - 5 \)
Since \( T_n = 6n - 5 \) is linear in \( n \), the common difference is \( d = 6 \) (constant), so the sequence is arithmetic.

3. Geometric Sequences

A geometric sequence is a list of numbers where you multiply by the same value each time. That value is the common ratio, \( r \).

The Key Idea

\( 3; \; 6; \; 12; \; 24; \; \ldots \) — here \( r = 2 \) (each term is doubled).

\( 81; \; 27; \; 9; \; 3; \; \ldots \) — here \( r = \frac{1}{3} \).

To find \( r \): divide any term by the one before it: \( r = \frac{T_2}{T_1} \).

The Formula

$$ T_n = a \cdot r^{n-1} $$

Notice: the exponent is \( n - 1 \), not \( n \). The first term \( T_1 = a \cdot r^0 = a \).

Confusing \( T_n = ar^{n-1} \) with \( T_n = ar^n \). The exponent is always one less than the term number.
Q7 Easy
Find the 8th term of: \( 2; \; 6; \; 18; \; 54; \; \ldots \)
Solution
\( a = 2, \quad r = \frac{6}{2} = 3, \quad n = 8 \)
\( T_8 = 2 \cdot 3^7 = 2 \times 2187 = 4374 \)
Q8 Medium
The 2nd term of a geometric sequence is 12 and the 5th term is 324. Find \( a \) and \( r \).
Solution
\( T_2 = ar = 12 \) ... (1)
\( T_5 = ar^4 = 324 \) ... (2)
Divide (2) by (1): \( \frac{ar^4}{ar} = \frac{324}{12} \)
\( r^3 = 27 \), so \( r = 3 \)
From (1): \( a(3) = 12 \), so \( a = 4 \)
Q9 Medium
If \( 3; \; x; \; 27 \) are three consecutive terms of a geometric sequence, find the value(s) of \( x \).
Solution
In a geometric sequence, consecutive terms have a constant ratio:
\( \frac{x}{3} = \frac{27}{x} \)
\( x^2 = 81 \)
\( x = \pm 9 \)

4. Geometric Series (Finite)

Adding up the terms of a geometric sequence gives a geometric series.

The Sum Formula

$$ S_n = \frac{a(r^n - 1)}{r - 1} \quad \text{where } r \neq 1 $$

An equivalent form (useful when \( |r| < 1 \)):

$$ S_n = \frac{a(1 - r^n)}{1 - r} $$

Worked Example

Find the sum of the first 6 terms of: \( 4; \; 12; \; 36; \; \ldots \)

\( a = 4, \quad r = 3, \quad n = 6 \)
\( S_6 = \frac{4(3^6 - 1)}{3 - 1} = \frac{4(729 - 1)}{2} = \frac{4 \times 728}{2} = 1456 \)
Q10 Easy
Find the sum of the first 8 terms of: \( 1; \; 2; \; 4; \; 8; \; \ldots \)
Solution
\( a = 1, \quad r = 2, \quad n = 8 \)
\( S_8 = \frac{1(2^8 - 1)}{2 - 1} = \frac{256 - 1}{1} = 255 \)
Q11 Medium
A geometric series has \( a = 5 \) and \( r = 2 \). If \( S_n = 635 \), find \( n \).
Solution
\( \frac{5(2^n - 1)}{2 - 1} = 635 \)
\( 5(2^n - 1) = 635 \)
\( 2^n - 1 = 127 \)
\( 2^n = 128 = 2^7 \)
\( n = 7 \)

5. Convergent Geometric Series

Here's where it gets interesting. If the common ratio \( r \) is between \( -1 \) and \( 1 \) (exclusive), the terms get smaller and smaller. The sum approaches a finite limit even if you add infinitely many terms.

Sum to Infinity

$$ S_\infty = \frac{a}{1 - r} \quad \text{only if } -1 < r < 1 $$

The series is said to converge. If \( |r| \geq 1 \), the series diverges (no finite sum).

Worked Example

Find \( S_\infty \) for: \( 18; \; 6; \; 2; \; \frac{2}{3}; \; \ldots \)

\( a = 18, \quad r = \frac{6}{18} = \frac{1}{3} \)
Since \( |r| = \frac{1}{3} < 1 \), the series converges.
\( S_\infty = \frac{18}{1 - \frac{1}{3}} = \frac{18}{\frac{2}{3}} = 27 \)
Always check that \( -1 < r < 1 \) before using \( S_\infty \). If \( |r| \geq 1 \), the sum to infinity does not exist.
Q12 Easy
Find \( S_\infty \) for: \( 8; \; 4; \; 2; \; 1; \; \ldots \)
Solution
\( a = 8, \quad r = \frac{1}{2} \). Since \( |r| < 1 \), the series converges.
\( S_\infty = \frac{8}{1 - \frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16 \)
Q13 Medium
A convergent geometric series has a first term of 10 and \( S_\infty = 40 \). Find \( r \).
Solution
\( \frac{10}{1 - r} = 40 \)
\( 10 = 40(1 - r) = 40 - 40r \)
\( 40r = 30 \)
\( r = \frac{3}{4} \)
Check: \( |r| = \frac{3}{4} < 1 \). Valid.
Q14 Hard
For which values of \( x \) will the series \( (x+1) + (x+1)^2 + (x+1)^3 + \ldots \) converge?
Solution
This is geometric with \( a = (x+1) \) and \( r = (x+1) \).
For convergence: \( -1 < r < 1 \), i.e. \( -1 < x + 1 < 1 \).
\( -2 < x < 0 \)
But also \( r \neq 0 \) (otherwise \( a = 0 \) and the series is trivially 0). So we also exclude \( x = -1 \).
Answer: \( x \in (-2, \; 0), \; x \neq -1 \)

6. Sigma Notation

Sigma notation (\( \sum \)) is a compact way to write a sum. The symbol \( \sum \) means "add up".

How to Read It

\( \displaystyle \sum_{k=1}^{5} (2k+1) \) means: substitute \( k = 1, 2, 3, 4, 5 \) into \( 2k + 1 \) and add them all up.

\( = 3 + 5 + 7 + 9 + 11 = 35 \)

Look at the general term to decide if it's arithmetic or geometric. If the term contains \( k \) (linear), it's arithmetic. If it contains \( r^k \), it's geometric.
Worked Example

Evaluate: \( \displaystyle \sum_{k=1}^{6} 3 \cdot 2^{k-1} \)

This is a geometric series with \( a = 3, \; r = 2, \; n = 6 \).
\( S_6 = \frac{3(2^6 - 1)}{2 - 1} = 3(63) = 189 \)
Q15 Easy
Expand and evaluate: \( \displaystyle \sum_{k=1}^{4} (3k - 1) \)
Solution
\( k=1: \; 3(1)-1 = 2 \)
\( k=2: \; 3(2)-1 = 5 \)
\( k=3: \; 3(3)-1 = 8 \)
\( k=4: \; 3(4)-1 = 11 \)
Sum \( = 2 + 5 + 8 + 11 = 26 \)
Q16 Medium
Write in sigma notation: \( 5 + 10 + 20 + 40 + \ldots + 5120 \)
Solution
Geometric with \( a = 5, \; r = 2 \). The general term is \( T_k = 5 \cdot 2^{k-1} \).
Find \( n \): \( 5 \cdot 2^{n-1} = 5120 \), so \( 2^{n-1} = 1024 = 2^{10} \), giving \( n = 11 \).
\( \displaystyle \sum_{k=1}^{11} 5 \cdot 2^{k-1} \)

7. Mixed Problems (IEB Level)

These combine multiple concepts — exactly what you'll see in the exam. Take your time.

Q17 Hard

The first three terms of an arithmetic sequence are \( 2p + 3; \; p + 6; \; p + 2 \).

(a) Find the value of \( p \).

(b) Find the sum of the first 25 terms.

Solution
(a) For an arithmetic sequence: \( T_2 - T_1 = T_3 - T_2 \)
\( (p + 6) - (2p + 3) = (p + 2) - (p + 6) \)
\( -p + 3 = -4 \)
\( p = 7 \)
Check: terms are \( 17; \; 13; \; 9 \) with \( d = -4 \). Correct.
(b) \( a = 17, \; d = -4, \; n = 25 \)
\( S_{25} = \frac{25}{2}[2(17) + 24(-4)] = \frac{25}{2}[34 - 96] = \frac{25}{2}(-62) = -775 \)
Q18 Hard

\( \displaystyle \sum_{k=1}^{\infty} 8 \left(\frac{1}{2}\right)^k = S_\infty \).

(a) Find \( S_\infty \).

(b) Find the smallest value of \( n \) for which \( S_\infty - S_n < 0.01 \).

Solution
(a) When \( k = 1 \): first term = \( 8 \cdot \frac{1}{2} = 4 \). So \( a = 4, \; r = \frac{1}{2} \).
\( S_\infty = \frac{4}{1 - \frac{1}{2}} = 8 \)
(b) \( S_\infty - S_n = 8 - \frac{4(1 - (1/2)^n)}{1/2} = 8 - 8(1 - (1/2)^n) = 8 \cdot (1/2)^n \)
We need \( 8 \cdot (1/2)^n < 0.01 \), so \( (1/2)^n < 0.00125 \).
\( 2^n > 800 \). Since \( 2^9 = 512 \) and \( 2^{10} = 1024 \), we need \( n \geq 10 \).
\( n = 10 \)
Q19 Hard

The sum of the first 20 terms of an arithmetic series is 610, and the sum of the first 10 terms is 155. Find the first term and the common difference.

Solution
\( S_{20} = 10(2a + 19d) = 610 \), so \( 2a + 19d = 61 \) ... (1)
\( S_{10} = 5(2a + 9d) = 155 \), so \( 2a + 9d = 31 \) ... (2)
Subtract (2) from (1): \( 10d = 30 \), so \( d = 3 \)
Substitute into (2): \( 2a + 27 = 31 \), so \( a = 2 \)
Check: \( S_{20} = 10[4 + 57] = 610 \). Correct.
Q20 Hard

Given the series \( \displaystyle \sum_{k=1}^{n} (5k - 3) \):

(a) Write down the first three terms.

(b) Find \( n \) if the sum equals 632.

Solution
(a) \( k=1: \; 5(1)-3 = 2 \); \( k=2: \; 5(2)-3 = 7 \); \( k=3: \; 5(3)-3 = 12 \)
First three terms: \( 2; \; 7; \; 12 \)
(b) Arithmetic with \( a = 2, \; d = 5 \).
\( S_n = \frac{n}{2}(5n - 1) = 632 \)
\( n(5n - 1) = 1264 \)
\( 5n^2 - n - 1264 = 0 \)
Using the quadratic formula: \( n = \frac{1 \pm \sqrt{1 + 25280}}{10} = \frac{1 \pm \sqrt{25281}}{10} = \frac{1 \pm 159}{10} \)
\( n = \frac{160}{10} = 16 \) (reject the negative root)
Check: \( S_{16} = \frac{16}{2}(5 \times 16 - 1) = 8 \times 79 = 632 \). Correct.