1. What is a Sequence?
Dear student, welcome to Unit 1 of your Grade 12 Mathematics! In this unit, we study Sequences and Series — one of the most beautiful and useful topics in mathematics.
A sequence is an ordered list of numbers that follows a specific rule or pattern. Each number in the sequence is called a term.
For example:
- \( 2, 5, 8, 11, 14, \ldots \) — each term increases by 3
- \( 3, 6, 12, 24, 48, \ldots \) — each term is multiplied by 2
- \( 1, 1, 2, 3, 5, 8, 13, \ldots \) — each term is the sum of the two previous terms (Fibonacci)
We denote the terms of a sequence as \( t_1, t_2, t_3, \ldots, t_n \) or as \( a_1, a_2, a_3, \ldots, a_n \). The subscript tells us the position of the term.
Can you see the pattern in each sequence above? Recognizing patterns is the key skill in this unit!
Sequence: \( 2, 5, 8, 11 \)
Series: \( 2 + 5 + 8 + 11 \)
1.1 Finite and Infinite Sequences
- Finite sequence: has a last term, e.g., \( 1, 4, 9, 16, 25 \) (5 terms)
- Infinite sequence: continues forever, indicated by \( \ldots \), e.g., \( 1, 2, 3, 4, \ldots \)
1.2 General Term (nth Term)
The general term or nth term of a sequence, written as \( t_n \) or \( a_n \), is a formula that gives us any term when we substitute its position number \( n \).
Worked Example 1: Find the first 4 terms and the 10th term of the sequence whose nth term is \( t_n = 3n – 1 \).
Solution:
\( t_1 = 3(1) – 1 = 2 \)
\( t_2 = 3(2) – 1 = 5 \)
\( t_3 = 3(3) – 1 = 8 \)
\( t_4 = 3(4) – 1 = 11 \)
First 4 terms: \( 2, 5, 8, 11 \)
\( t_{10} = 3(10) – 1 = 29 \)
Worked Example 2: Find the nth term of the sequence: \( 1, 4, 9, 16, 25, \ldots \)
Solution:
Position: \( 1, 2, 3, 4, 5 \)
Term: \( 1, 4, 9, 16, 25 \)
Notice: \( 1 = 1^2, \; 4 = 2^2, \; 9 = 3^2, \; 16 = 4^2, \; 25 = 5^2 \)
Worked Example 3: Find the nth term of: \( 2, 5, 10, 17, 26, \ldots \)
Solution:
Differences between consecutive terms: \( 3, 5, 7, 9, \ldots \) (odd numbers, increasing by 2)
This suggests \( t_n = n^2 + 1 \). Let’s check:
\( t_1 = 1 + 1 = 2 \) ✓, \( t_2 = 4 + 1 = 5 \) ✓, \( t_3 = 9 + 1 = 10 \) ✓, \( t_4 = 16 + 1 = 17 \) ✓
Find the nth term of the sequence: \( 3, 7, 11, 15, 19, \ldots \) Hence find the 20th term.
Differences: \( 4, 4, 4, 4 \) — constant difference of 4.
The pattern: \( t_n = 4n – 1 \). Check: \( t_1 = 3 \) ✓, \( t_2 = 7 \) ✓, \( t_3 = 11 \) ✓
Find the nth term of: \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots \)
Numerator: \( 1, 2, 3, 4 \) → \( n \)
Denominator: \( 2, 3, 4, 5 \) → \( n + 1 \)
2. Arithmetic Sequences (Arithmetic Progression — AP)
An Arithmetic Sequence (or Arithmetic Progression, abbreviated AP) is a sequence where the difference between any two consecutive terms is always the same. This constant difference is called the common difference, denoted by \( d \).
Think of it like walking up stairs where every step is the same height. If you go up by 3 each time: 5, 8, 11, 14, 17, … that’s an AP with \( d = 3 \).
2.1 The General Term of an AP
- \( a_n \) = the \( n \)-th term
- \( a_1 \) = the first term
- \( d \) = the common difference
- \( n \) = the position number
Worked Example 4: In the AP \( 7, 11, 15, 19, \ldots \), find: (a) the common difference, (b) the 25th term, (c) which term equals 127.
Solution:
(a) \( d = 11 – 7 = 4 \)
(b) \( a_{25} = 7 + (25 – 1) \times 4 = 7 + 96 = 103 \)
(c) Set \( a_n = 127 \):
So the 31st term is 127.
Worked Example 5: The 8th term of an AP is 37 and the 15th term is 65. Find the first term and common difference.
Solution:
Subtract (i) from (ii):
Substitute in (i): \( a_1 + 28 = 37 \), so \( a_1 = 9 \)
The AP is \( 9, 13, 17, 21, \ldots \)
Worked Example 6: The 5th term of an AP is twice the 2nd term, and the 9th term exceeds the 4th term by 15. Find the AP.
Solution:
From (i): \( a_1 = 2(3) = 6 \)
The AP is \( 6, 9, 12, 15, 18, \ldots \)
- Always find \( d \) first by subtracting any two consecutive terms.
- If given two terms (e.g., \( a_m \) and \( a_n \)), set up two equations and solve simultaneously.
- The formula \( a_n = a_1 + (n-1)d \) works for any term — you do NOT need to list all terms.
- Common mistake: using \( n \) instead of \( n-1 \). Remember: for the 1st term, \( n-1 = 0 \), so \( a_1 = a_1 + 0 \). This is correct!
An AP has first term \( -8 \) and common difference 3. (a) Find the 15th term. (b) Is 100 a term of this AP? If so, which term?
(a) \( a_{15} = -8 + (15-1) \times 3 = -8 + 42 = 34 \)
(b) Set \( a_n = 100 \):
Yes, 100 is the 37th term of this AP.
If the 3rd and 7th terms of an AP are 14 and 30 respectively, find the first term, the common difference, and the 20th term.
Subtract (i) from (ii): \( 4d = 16 \), so \( d = 4 \)
From (i): \( a_1 + 8 = 14 \), so \( a_1 = 6 \)
3. Arithmetic Mean
If three numbers are in AP, the middle number is called the Arithmetic Mean (AM) of the other two.
Worked Example 7: Insert 3 arithmetic means between 4 and 24.
Solution: We need 5 numbers in total in AP: \( 4, \_, \_, \_, 24 \)
Here \( a_1 = 4 \), \( a_5 = 24 \), \( n = 5 \)
The AP is: \( 4, 9, 14, 19, 24 \)
The three arithmetic means are: \( 9, 14, 19 \)
Insert 4 arithmetic means between 3 and 23.
Total terms = 6. \( a_1 = 3 \), \( a_6 = 23 \)
AP: \( 3, 7, 11, 15, 19, 23 \)
Four arithmetic means: \( 7, 11, 15, 19 \)
4. Arithmetic Series
An Arithmetic Series is the sum of the terms of an arithmetic sequence.
4.1 Sum of an Arithmetic Series
This formula makes sense! If you add the first and last term, you get \( a_1 + a_n \). If you add the second and second-to-last, you get the same total \( a_2 + a_{n-1} = a_1 + a_n \). There are \( n/2 \) such pairs (or \( n \) terms with average \( (a_1+a_n)/2 \)).
Worked Example 8: Find the sum of the first 20 terms of the AP: \( 3, 7, 11, 15, \ldots \)
Solution:
\( a_1 = 3 \), \( d = 4 \), \( n = 20 \)
Worked Example 9: Find the sum of all positive integers from 1 to 100.
Solution:
\( a_1 = 1 \), \( a_{100} = 100 \), \( n = 100 \)
This famous result was reportedly found by the mathematician Gauss as a young student!
Worked Example 10: How many terms of the AP \( 18, 15, 12, 9, \ldots \) are needed to give a sum of 0?
Solution:
\( a_1 = 18 \), \( d = -3 \) (decreasing AP!), \( S_n = 0 \)
So \( n = 0 \) (reject) or \( 39 – 3n = 0 \), giving \( n = 13 \)
- For a decreasing AP, \( d \) is negative. The formulas still work perfectly!
- If \( S_n \) is given and you need to find \( n \), you will get a quadratic equation. Reject any negative or zero solution for \( n \).
- Use \( S_n = \frac{n}{2}(a_1 + a_n) \) when you know the last term. Use \( S_n = \frac{n}{2}[2a_1 + (n-1)d] \) when you don’t.
The sum of the first 15 terms of an AP is 525. The first term is 8. Find the common difference and the 15th term.
5. Sigma Notation
Sigma notation (written as \( \sum \), the Greek capital letter “sigma”) is a compact way to write the sum of a sequence.
Worked Example 11: Evaluate \( \sum_{k=1}^{5} (3k + 1) \)
Solution:
Substitute \( k = 1, 2, 3, 4, 5 \):
Worked Example 12: Evaluate \( \sum_{k=1}^{20} (2k – 1) \)
Solution: This is the sum of the first 20 odd numbers: \( 1 + 3 + 5 + \cdots + 39 \)
This is an AP with \( a_1 = 1 \), \( d = 2 \), \( n = 20 \)
Worked Example 13: Evaluate \( \sum_{k=1}^{50} k^2 \)
Solution: This requires the sum of squares formula:
Evaluate \( \sum_{k=1}^{100} k \) and \( \sum_{k=1}^{10} k^3 \).
Interesting fact: \( 1^3 + 2^3 + \cdots + 10^3 = (1+2+\cdots+10)^2 = 55^2 \). The sum of cubes always equals the square of the sum!
6. Geometric Sequences (Geometric Progression — GP)
A Geometric Sequence (or Geometric Progression, GP) is a sequence where the ratio of any two consecutive terms is always the same. This constant ratio is called the common ratio, denoted by \( r \).
Unlike an AP where we ADD a constant, in a GP we MULTIPLY by a constant. If \( r = 2 \), each term doubles: 3, 6, 12, 24, 48, … If \( r = \frac{1}{2} \), each term halves: 32, 16, 8, 4, 2, …
6.1 The General Term of a GP
- \( a_n \) = the \( n \)-th term
- \( a_1 \) = the first term
- \( r \) = the common ratio
- \( n \) = the position number
Worked Example 14: In the GP \( 5, 15, 45, 135, \ldots \), find: (a) the common ratio, (b) the 8th term, (c) which term equals \( 5 \times 3^{10} \).
Solution:
(a) \( r = \frac{15}{5} = 3 \)
(b) \( a_8 = 5 \times 3^{7} = 5 \times 2187 = 10935 \)
(c) Set \( a_n = 5 \times 3^{10} \):
Worked Example 15: The 4th term of a GP is 54 and the 7th term is 1458. Find the first term and common ratio.
Solution:
Divide (ii) by (i):
From (i): \( a_1 \times 27 = 54 \), so \( a_1 = 2 \)
The GP is \( 2, 6, 18, 54, \ldots \)
Worked Example 16: Find the 5th term of the GP: \( \frac{2}{3}, \frac{4}{9}, \frac{8}{27}, \ldots \)
Solution:
\( a_1 = \frac{2}{3} \), \( r = \frac{4/9}{2/3} = \frac{4}{9} \times \frac{3}{2} = \frac{2}{3} \)
- \( r \) can be a fraction (less than 1), which gives a decreasing GP.
- \( r \) can be negative, which gives an alternating GP: \( 3, -6, 12, -24, \ldots \)
- When given two terms of a GP, divide the equations to eliminate \( a_1 \) and find \( r \) first.
- Be careful with exponents: \( a_n = a_1 \cdot r^{n-1} \) (power is \( n-1 \), not \( n \)).
A GP has first term 3 and common ratio \( \frac{1}{2} \). (a) Find the 10th term. (b) Which term is the first to be less than 0.01?
(a) \( a_{10} = 3 \times \left(\frac{1}{2}\right)^9 = 3 \times \frac{1}{512} = \frac{3}{512} \approx 0.00586 \)
(b) We need \( a_n < 0.01 \):
\( 2^8 = 256 \) (not > 300), \( 2^9 = 512 > 300 \). So \( n – 1 = 9 \), \( n = 10 \).
The 10th term is the first term less than 0.01.
7. Geometric Mean
If three positive numbers are in GP, the middle number is called the Geometric Mean (GM) of the other two.
Worked Example 17: Insert 2 geometric means between 4 and 108.
Solution: We need 4 numbers in GP: \( 4, \_, \_, 108 \)
\( a_1 = 4 \), \( a_4 = 108 \), \( n = 4 \)
The GP is: \( 4, 12, 36, 108 \)
The two geometric means are: \( 12, 36 \)
Insert 3 geometric means between 2 and 162.
Total terms = 5. \( a_1 = 2 \), \( a_5 = 162 \)
GP: \( 2, 6, 18, 54, 162 \)
Three geometric means: \( 6, 18, 54 \)
8. Geometric Series
A Geometric Series is the sum of the terms of a geometric sequence.
8.1 Sum of a Finite Geometric Series
Worked Example 18: Find the sum of the first 8 terms of the GP: \( 3, 6, 12, 24, \ldots \)
Solution:
\( a_1 = 3 \), \( r = 2 \), \( n = 8 \)
Worked Example 19: Find the sum: \( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{256} \)
Solution:
\( a_1 = 1 \), \( r = \frac{1}{2} \). We need to find \( n \). \( \frac{1}{256} = \left(\frac{1}{2}\right)^8 \), so this is \( a_9 \), meaning \( n = 9 \).
8.2 Infinite Geometric Series
When \( |r| < 1 \) (i.e., \( -1 < r < 1 \)), the terms get smaller and smaller, and the series approaches a finite value. We can find the sum to infinity:
Think about it: if you keep adding half of what’s left (\( r = \frac{1}{2} \)), you keep getting closer to 2 but never exceed it. That’s why the sum converges!
Worked Example 20: Find the sum to infinity: \( 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \)
Solution:
\( a_1 = 1 \), \( r = \frac{1}{3} \). Since \( |r| = \frac{1}{3} < 1 \), the sum converges.
Worked Example 21: Express \( 0.\overline{7} \) (meaning \( 0.7777\ldots \)) as a fraction.
Solution:
This is an infinite GP with \( a_1 = 0.7 \), \( r = 0.1 \)
Worked Example 22: Express \( 0.\overline{63} \) (meaning \( 0.636363\ldots \)) as a fraction.
Solution:
\( a_1 = 0.63 \), \( r = 0.01 \)
- \( S_{\infty} \) ONLY exists when \( |r| < 1 \). If \( |r| \geq 1 \), the sum diverges (goes to infinity).
- For converting repeating decimals: the repeating part is \( a_1 \), and \( r = \frac{1}{10^k} \) where \( k \) is the number of repeating digits.
- Use \( S_n = \frac{a_1(1-r^n)}{1-r} \) when \( |r| < 1 \) (avoids negative numerator issues).
- Use \( S_n = \frac{a_1(r^n – 1)}{r-1} \) when \( r > 1 \).
(a) Find the sum to infinity of \( 8 + 4 + 2 + 1 + \cdots \)
(b) Express \( 0.\overline{45} \) as a fraction in lowest terms.
(a) \( a_1 = 8 \), \( r = \frac{1}{2} \). Since \( |r| < 1 \):
(b) \( 0.\overline{45} = 0.45 + 0.0045 + 0.000045 + \cdots \)
\( a_1 = 0.45 \), \( r = 0.01 \)
9. Summary of Key Exam Notes
- AP: constant difference \( d \). Formula: \( a_n = a_1 + (n-1)d \). Sum: \( S_n = \frac{n}{2}(a_1 + a_n) \)
- GP: constant ratio \( r \). Formula: \( a_n = a_1 \cdot r^{n-1} \). Sum: \( S_n = \frac{a_1(r^n-1)}{r-1} \)
- Infinite GP: \( S_{\infty} = \frac{a_1}{1-r} \), only when \( |r| < 1 \)
- Arithmetic Mean: \( A = \frac{a+b}{2} \)
- Geometric Mean: \( G = \sqrt{ab} \) (for positive numbers)
- Sigma formulas: \( \sum k = \frac{n(n+1)}{2} \), \( \sum k^2 = \frac{n(n+1)(2n+1)}{6} \), \( \sum k^3 = \left[\frac{n(n+1)}{2}\right]^2 \)
- Repeating decimals → fractions: use infinite GP with \( r = 10^{-k} \)
- Always check: is the sequence AP (constant difference) or GP (constant ratio)?
- For decreasing sequences, \( d < 0 \) (AP) or \( 0 < r < 1 \) (GP).
An AP has first term 5 and common difference 3. A GP has first term 5 and common ratio 2. After how many terms will the GP first exceed the sum of the first \( n \) terms of the AP?
AP sum: \( S_n = \frac{n}{2}[2(5) + (n-1)(3)] = \frac{n}{2}(10 + 3n – 3) = \frac{n(3n+7)}{2} \)
GP \( n \)-th term: \( a_n = 5 \times 2^{n-1} \)
We need \( 5 \times 2^{n-1} > \frac{n(3n+7)}{2} \), i.e., \( 10 \times 2^{n-1} > n(3n+7) \)
Test values:
| \( n \) | GP term \( 5 \times 2^{n-1} \) | AP sum \( \frac{n(3n+7)}{2} \) | GP > AP sum? |
|---|---|---|---|
| 1 | 5 | 5 | No |
| 2 | 10 | 13 | No |
| 3 | 20 | 24 | No |
| 4 | 40 | 38 | Yes ✓ |
After 4 terms, the GP term first exceeds the AP sum.
Quick Revision Notes — Sequences and Series
1. Important Definitions
- Sequence: An ordered list of numbers following a pattern (e.g., \( 2, 5, 8, \ldots \))
- Series: The sum of the terms of a sequence (e.g., \( 2 + 5 + 8 + \cdots \))
- nth term (\( a_n \)): A formula giving any term by its position
- Arithmetic Progression (AP): Sequence with constant difference \( d \) between consecutive terms
- Geometric Progression (GP): Sequence with constant ratio \( r \) between consecutive terms
- Arithmetic Mean (AM): \( \frac{a+b}{2} \) — middle term when three numbers are in AP
- Geometric Mean (GM): \( \sqrt{ab} \) — middle term when three positive numbers are in GP
- Sigma notation (\( \sum \)): Compact notation for writing sums
- Infinite series: Sum of infinitely many terms (only converges for GP when \( |r| < 1 \))
2. All Key Formulas
ARITHMETIC PROGRESSION:
GEOMETRIC PROGRESSION:
SIGMA NOTATION:
REPEATING DECIMALS:
Example: \( 0.\overline{123} = \frac{123}{999} = \frac{41}{333} \)
3. Quick Identification Guide
| Feature | Arithmetic (AP) | Geometric (GP) |
|---|---|---|
| Pattern | ADD constant \( d \) | MULTIPLY by constant \( r \) |
| How to check | \( a_2 – a_1 = a_3 – a_2 = \cdots \) | \( \frac{a_2}{a_1} = \frac{a_3}{a_2} = \cdots \) |
| nth term | \( a_1 + (n-1)d \) | \( a_1 \cdot r^{n-1} \) |
| Sum | \( \frac{n}{2}(a_1 + a_n) \) | \( \frac{a_1(r^n-1)}{r-1} \) |
| Mean | \( \frac{a+b}{2} \) | \( \sqrt{ab} \) |
| Infinite sum | Never converges (unless \( d=0 \)) | \( \frac{a_1}{1-r} \) if \( |r|<1 \) |
4. Common Mistakes to Avoid
- Using \( n \) instead of \( n-1 \): The nth term is \( a_1 + (n-1)d \), NOT \( a_1 + nd \). Check with \( n=1 \): \( a_1 + 0 = a_1 \) ✓
- Confusing AP and GP: Always check: is it addition (AP) or multiplication (GP)?
- Forgetting \( |r| < 1 \) condition: You can ONLY use \( S_{\infty} = \frac{a_1}{1-r} \) when \( |r| < 1 \). If \( r = 3 \), the infinite sum does NOT exist!
- Wrong formula for \( S_n \) when \( r < 1 \): Use \( \frac{a_1(1-r^n)}{1-r} \) to avoid negative values.
- Negative \( r \): A GP can have \( r < 0 \), giving alternating signs. The formulas still work!
- Repeating decimal error: For \( 0.\overline{3} \), \( a_1 = 0.3 \) (not 3) and \( r = 0.1 \).
- Arithmetic vs geometric mean: AM uses addition (average); GM uses multiplication (square root of product).
- Inserting means: “Insert 3 arithmetic means between \( a \) and \( b \)” means the total has 5 terms, NOT 3.
- Quadratic equations for \( n \): When finding \( n \) from \( S_n \), you get a quadratic. Reject negative or zero \( n \).
- Sigma limits: \( \sum_{k=1}^{n} \) means starting from \( k=1 \). If it says \( \sum_{k=0}^{n} \), the first term uses \( k=0 \).
5. Quick Examples
Q: AP: \( a_1 = 10 \), \( d = -3 \). Find \( a_{12} \).
A: \( a_{12} = 10 + 11(-3) = 10 – 33 = -23 \)
Q: GP: \( a_1 = 2 \), \( r = 3 \). Find \( S_5 \).
A: \( S_5 = \frac{2(3^5 – 1)}{3-1} = \frac{2(243-1)}{2} = 242 \)
Q: \( S_{\infty} \) of \( 6 + 2 + \frac{2}{3} + \cdots \)?
A: \( r = \frac{1}{3} \). \( S_{\infty} = \frac{6}{1-1/3} = \frac{6}{2/3} = 9 \)
Q: \( 0.\overline{8} \) as a fraction?
A: \( \frac{0.8}{1-0.1} = \frac{0.8}{0.9} = \frac{8}{9} \)
Q: GM of 4 and 9?
A: \( \sqrt{4 \times 9} = \sqrt{36} = 6 \)
Challenge Exam Questions — Sequences and Series
These questions test your deep understanding. Try each one fully before checking the answer!
Section A: Multiple Choice Questions
The 10th term of the AP \( 4, 7, 10, 13, \ldots \) is:
A) 29 B) 31 C) 34 D) 30
Answer: B) 31
The sum to infinity of \( 5 – \frac{5}{2} + \frac{5}{4} – \frac{5}{8} + \cdots \) is:
A) 10 B) \( \frac{10}{3} \) C) \( \frac{5}{3} \) D) Does not exist
Answer: B) \( \frac{10}{3} \)
\( a_1 = 5 \), \( r = -\frac{1}{2} \). Since \( |r| = \frac{1}{2} < 1 \):
Note: Negative \( r \) is fine as long as \( |r| < 1 \)!
If \( 2x, x+10, 3x+2 \) are three consecutive terms of an AP, then \( x = \):
A) 4 B) 6 C) 8 D) 3
Answer: B) 6
For three terms in AP: \( 2 \times \text{middle} = \text{first} + \text{third} \)
The sum \( \sum_{k=1}^{6} (k^2 + 1) \) equals:
A) 91 B) 97 C) 86 D) 92
Answer: B) 97
The geometric mean of \( \frac{1}{2} \) and \( 8 \) is:
A) 4 B) 2 C) \( \frac{17}{4} \) D) \( \sqrt{\frac{17}{2}} \)
Answer: B) 2
Section B: Fill in the Blanks
In an AP, if \( a_1 = 3 \) and \( a_{10} = 30 \), then \( d = \) ________
Answer: 3
The 6th term of the GP \( 1, -2, 4, -8, \ldots \) is ________
Answer: 32
Wait — let me recheck: \( a_1 = 1, r = -2, a_6 = 1 \times (-2)^5 = -32 \). The answer is \(-32\).
\( 0.\overline{27} = \) ________ (as a fraction in lowest terms)
Answer: \( \frac{3}{11} \)
If \( \sum_{k=1}^{n} k = 55 \), then \( n = \) ________
Answer: 10
The number of terms in the AP \( 7, 11, 15, \ldots, 79 \) is ________
Answer: 19
Section C: Short Answer Questions
How can you quickly check if three numbers are in AP? How about GP?
AP check: Three numbers \( a, b, c \) are in AP if and only if \( 2b = a + c \) (i.e., twice the middle equals the sum of the extremes). Equivalently, \( b – a = c – b \).
GP check: Three positive numbers \( a, b, c \) are in GP if and only if \( b^2 = ac \) (i.e., the middle squared equals the product of the extremes). Equivalently, \( \frac{b}{a} = \frac{c}{b} \).
Why does the infinite geometric series \( 1 + 2 + 4 + 8 + \cdots \) NOT have a finite sum?
The common ratio is \( r = 2 \). Since \( |r| = 2 > 1 \), the terms grow larger and larger instead of approaching zero. The partial sums are \( 1, 3, 7, 15, 31, \ldots \) which grow without bound. The infinite sum formula \( S_{\infty} = \frac{a_1}{1-r} \) is ONLY valid when \( |r| < 1 \).
Find the sum of the first 100 odd numbers without adding them one by one.
The first 100 odd numbers form an AP: \( 1, 3, 5, \ldots \) with \( a_1 = 1 \), \( d = 2 \), \( n = 100 \).
The 100th odd number: \( a_{100} = 1 + 99 \times 2 = 199 \)
The sum of the first \( n \) odd numbers is always \( n^2 \). Here \( 100^2 = 10000 \). ✓
Section D: Step-by-Step Calculation Questions
The sum of the first 8 terms of an AP is 124. The sum of the first 16 terms is 448. Find the first term and common difference.
From (i): \( 2a_1 + 7d = 31 \) … (iii)
From (ii): \( 2a_1 + 15d = 56 \) … (iv)
Subtract (iii) from (iv): \( 8d = 25 \), so \( d = \frac{25}{8} = 3.125 \)
From (iii): \( 2a_1 + 7 \times \frac{25}{8} = 31 \), \( 2a_1 = 31 – \frac{175}{8} = \frac{248-175}{8} = \frac{73}{8} \)
A ball is dropped from a height of 80 meters. Each time it bounces, it rises to 75% of its previous height. Find the total distance travelled by the ball before it comes to rest.
The ball falls 80 m, then rises \( 80 \times 0.75 \), then falls \( 80 \times 0.75 \), then rises \( 80 \times 0.75^2 \), etc.
The infinite GP inside the brackets: \( a_1 = 60 \), \( r = 0.75 \)
Key insight: After the first fall, every bounce involves going UP and DOWN the same distance, so we multiply by 2!
Evaluate \( \sum_{k=1}^{20} (2k + 3) – \sum_{k=1}^{15} (3k – 1) \)
The 4th term of a GP is 2 and the 7th term is \( \frac{16}{27} \). Find the sum of the first 10 terms.
Divide (ii) by (i): \( r^3 = \frac{16/27}{2} = \frac{8}{27} \), so \( r = \frac{2}{3} \)
From (i): \( a_1 \times \frac{8}{27} = 2 \), so \( a_1 = \frac{27}{4} \)
The sum of three numbers in AP is 27 and their product is 504. Find the three numbers.
Let the three numbers be \( a-d \), \( a \), \( a+d \).
When \( d = 5 \): \( 4, 9, 14 \). When \( d = -5 \): \( 14, 9, 4 \). Same set!
If \( S_n = 3n^2 + 2n \) is the sum of the first \( n \) terms of an AP, find: (a) the first term, (b) the common difference, (c) the 20th term.
(a) \( a_1 = S_1 = 3(1)^2 + 2(1) = 5 \)
(b) \( S_2 = 3(4) + 4 = 16 \). So \( a_2 = S_2 – S_1 = 16 – 5 = 11 \). Thus \( d = a_2 – a_1 = 6 \).
Alternative method: \( S_n = \frac{n}{2}[2a_1 + (n-1)d] \). Comparing with \( 3n^2 + 2n = \frac{n}{2}(6n + 4) = \frac{n}{2}[6 + 6n – 6] = \frac{n}{2}[2(3) + (n-1)(6)] \). So \( a_1 = 3 \)… Wait, let me redo.
This gives \( a_1 = 2 \) and \( d = 6 \). But \( S_1 = 3 + 2 = 5 \), not 2. The formula comparison method is tricky. Let me use the direct approach:
\( a_1 = S_1 = 5 \), \( a_2 = S_2 – S_1 = 16 – 5 = 11 \), \( d = 6 \).
Answer: \( a_1 = 5 \), \( d = 6 \), \( a_{20} = 119 \)
Express \( 0.2\overline{17} \) (meaning \( 0.2171717\ldots \)) as a fraction.
The repeating part is an infinite GP: \( a_1 = 0.017 \), \( r = 0.01 \)
Shortcut formula: \( 0.2\overline{17} = \frac{217 – 2}{990} = \frac{215}{990} = \frac{43}{198} \)
Prove that the sum of the first \( n \) terms of the series \( 1 + 2 + 4 + 7 + 11 + \cdots \) is \( \frac{n(n^2 + 1)}{2} \). (Hint: the differences of consecutive terms form an AP.)
The given series: \( 1, 2, 4, 7, 11, \ldots \)
First differences: \( 1, 2, 3, 4, \ldots \) — this is an AP with \( d = 1 \)!
Such a sequence is called a “quadratic sequence.” The nth term is:
Now find \( S_n = \sum_{k=1}^{n} t_k \):
Wait, this gives \( \frac{n(n^2+5)}{6} \), not \( \frac{n(n^2+1)}{2} \). Let me verify with small \( n \):
\( n = 1 \): sum = 1. \( \frac{1(1+1)}{2} = 1 \) ✓. \( \frac{1(1+5)}{6} = 1 \) ✓. Both give 1.
\( n = 2 \): sum = 1+2 = 3. \( \frac{2(8+1)}{2} = 9 \) ✗. So the formula \( \frac{n(n^2+1)}{2} \) is WRONG for this sequence!
The correct sum is \( \frac{n(n^2+5)}{6} \). The question’s stated formula is incorrect. The proof shows the actual answer.
A and B are two positive numbers. Their arithmetic mean is 25 and their geometric mean is 20. Find A and B.
From (i): \( B = 50 – A \). Substitute in (ii):
\( A = 40 \) or \( A = 10 \). So the numbers are \( 10 \) and \( 40 \).
Find the sum of all integers between 100 and 500 that are divisible by 7.
The smallest integer > 100 divisible by 7: \( 105 \) (since \( 100 \div 7 = 14.28 \), next is \( 15 \times 7 = 105 \))
The largest integer < 500 divisible by 7: \( 497 \) (since \( 500 \div 7 = 71.42 \), previous is \( 71 \times 7 = 497 \))
This is an AP: \( 105, 112, 119, \ldots, 497 \) with \( d = 7 \)