IB Mathematics AI – Topic 1

General Form:

\[ a, \quad a+d, \quad a+2d, \quad a+3d, \quad \ldots \]

where a is the first term and d is the common difference

Common Difference:

\[ d = u_{n+1} - u_n \]

nth Term Formula:

\[ u_n = a + (n-1)d \]

where \(u_n\) is the nth term, a is the first term, d is the common difference, n is the term number

Examples:

• 2, 5, 8, 11, 14, ... (a = 2, d = 3)
• 10, 7, 4, 1, -2, ... (a = 10, d = -3)
• 0.5, 1.0, 1.5, 2.0, ... (a = 0.5, d = 0.5)

⚠️ Common Pitfalls & Tips:

• Always identify 'a' (first term) and 'd' (common difference) first
• The coefficient of n is NOT the common difference – it's (n-1)d in the formula
• Negative common difference means sequence is decreasing
• To find d: subtract any term from the next term

Solution:

Given: a = 7, d = 8

(a) First four terms:

\(u_1 = 7\)

\(u_2 = 7 + 8 = 15\)

\(u_3 = 15 + 8 = 23\)

\(u_4 = 23 + 8 = 31\)

Answer: 7, 15, 23, 31

(b) Find 10th term:

Using \(u_n = a + (n-1)d\)

\[ u_{10} = 7 + (10-1)(8) = 7 + 9(8) = 7 + 72 = 79 \]

Answer: 79

(c) Find n when \(u_n = 199\):

Using \(u_n = a + (n-1)d\)

\[ 199 = 7 + (n-1)(8) \]

\[ 192 = (n-1)(8) \]

\[ n-1 = 24 \]

\[ n = 25 \]

Answer: n = 25 (the 25th term)

Sum of First n Terms:

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

Or equivalently:

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

where l is the last term, \(l = u_n = a + (n-1)d\)

Which formula to use?

• Use \(S_n = \frac{n}{2}(2a + (n-1)d)\) when you know a, n, and d
• Use \(S_n = \frac{n}{2}(a + l)\) when you know first term, last term, and n

⚠️ Common Pitfalls & Tips:

• Don't forget the \(\frac{n}{2}\) factor – this is NOT optional!
• Be careful with brackets: \((2a + (n-1)d)\) not \(2a + n-1d\)
• If you know the last term, the second formula is usually quicker
• Check your answer: \(S_n\) should be larger than individual terms for positive sequences

Solution:

Step 1: Identify a and d

First term: a = 3

Common difference: d = 7 - 3 = 4

Number of terms: n = 20

Step 2: Use sum formula

Using \(S_n = \frac{n}{2}(2a + (n-1)d)\)

\[ S_{20} = \frac{20}{2}(2(3) + (20-1)(4)) \]

\[ = 10(6 + 19 \times 4) \]

\[ = 10(6 + 76) \]

\[ = 10(82) = 820 \]

Answer: 820

General Form:

\[ a, \quad ar, \quad ar^2, \quad ar^3, \quad \ldots \]

Common Ratio:

\[ r = \frac{u_{n+1}}{u_n} \]

nth Term Formula:

\[ u_n = ar^{n-1} \]

where \(u_n\) is the nth term, a is the first term, r is the common ratio

Examples:

• 2, 6, 18, 54, ... (a = 2, r = 3)
• 100, 50, 25, 12.5, ... (a = 100, r = 0.5)
• 1, -2, 4, -8, ... (a = 1, r = -2)

⚠️ Common Pitfalls & Tips:

• The power is (n-1), NOT n: \(ar^{n-1}\)
• To find r: divide any term by the previous term
• r can be negative, giving alternating signs
• If |r| > 1, sequence grows rapidly; if |r| < 1, sequence approaches zero

Solution:

Given: a = 5, r = 2

(a) Find 6th term:

Using \(u_n = ar^{n-1}\)

\[ u_6 = 5 \times 2^{6-1} = 5 \times 2^5 = 5 \times 32 = 160 \]

Answer: 160

(b) Find n when \(u_n = 640\):

Using \(u_n = ar^{n-1}\)

\[ 640 = 5 \times 2^{n-1} \]

\[ 128 = 2^{n-1} \]

\[ 2^7 = 2^{n-1} \]

\[ n - 1 = 7 \]

\[ n = 8 \]

Answer: n = 8

Sum of First n Terms (r ≠ 1):

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

Both forms are equivalent; use first when |r| < 1, second when r > 1

Infinite Geometric Series (|r| < 1):

When |r| < 1, the series converges to a finite sum:

\[ S_{\infty} = \frac{a}{1-r} \quad \text{for } |r| < 1 \]

Convergence Conditions:

• Converges: |r| < 1 (series has finite sum)
• Diverges: |r| ≥ 1 (series does not have finite sum)

⚠️ Common Pitfalls & Tips:

• Infinite sum ONLY exists when |r| < 1
• Don't confuse \(S_n\) (finite sum) with \(S_{\infty}\) (infinite sum)
• Check |r| < 1 before using infinite sum formula
• For \(S_{\infty}\), there is NO 'n' in the formula

Solution:

(a) Common ratio:

First term: a = 12

\[ r = \frac{6}{12} = \frac{1}{2} = 0.5 \]

Answer: r = 0.5

(b) Sum of first 8 terms:

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

\[ S_8 = \frac{12(1-0.5^8)}{1-0.5} \]

\[ = \frac{12(1-0.00390625)}{0.5} \]

\[ = \frac{12 \times 0.99609375}{0.5} = \frac{11.953125}{0.5} = 23.90625 \]

Answer: 23.9 (3 s.f.)

(c) Sum to infinity:

Since |r| = 0.5 < 1, infinite sum exists

Using \(S_{\infty} = \frac{a}{1-r}\)

\[ S_{\infty} = \frac{12}{1-0.5} = \frac{12}{0.5} = 24 \]

Answer: 24

General Form:

\[ \sum_{k=1}^{n} u_k = u_1 + u_2 + u_3 + \cdots + u_n \]

where k is the index, starting value is below Σ, ending value is above Σ

Components:

• Σ (Sigma): Summation symbol
• k = 1: Starting index value
• n: Ending index value
• \(u_k\): General term expression

Examples:

\(\sum_{k=1}^{5} k = 1 + 2 + 3 + 4 + 5 = 15\)

\(\sum_{k=1}^{4} 2k = 2(1) + 2(2) + 2(3) + 2(4) = 2 + 4 + 6 + 8 = 20\)

\(\sum_{k=1}^{3} k^2 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14\)

For Arithmetic Series:

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

For Geometric Series:

\[ \sum_{k=1}^{n} ar^{k-1} = S_n = \frac{a(1-r^n)}{1-r} \]

⚠️ Common Pitfalls & Tips:

• The index variable (k, i, n) is just a placeholder – can be any letter
• Always substitute the starting value first to check the pattern
• Number of terms = (upper limit - lower limit + 1)
• Can use GDC to evaluate sigma notation directly

Solution:

Method 1: Recognize as arithmetic series

First, write out first few terms to identify pattern:

When k = 1: 3(1) + 2 = 5

When k = 2: 3(2) + 2 = 8

When k = 3: 3(3) + 2 = 11

Pattern: 5, 8, 11, 14, ...

This is arithmetic with a = 5, d = 3, n = 20

Using \(S_n = \frac{n}{2}(2a + (n-1)d)\)

\[ S_{20} = \frac{20}{2}(2(5) + (20-1)(3)) \]

\[ = 10(10 + 19 \times 3) \]

\[ = 10(10 + 57) = 10(67) = 670 \]

Method 2: Using sigma properties

\[ \sum_{k=1}^{20} (3k + 2) = \sum_{k=1}^{20} 3k + \sum_{k=1}^{20} 2 \]

\[ = 3\sum_{k=1}^{20} k + 2(20) \]

Using \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)

\[ = 3 \times \frac{20(21)}{2} + 40 = 3(210) + 40 = 630 + 40 = 670 \]

Answer: 670

Arithmetic

• \(u_n = a + (n-1)d\)
• \(S_n = \frac{n}{2}(2a + (n-1)d)\)
• or \(S_n = \frac{n}{2}(a + l)\)

Geometric

• \(u_n = ar^{n-1}\)
• \(S_n = \frac{a(1-r^n)}{1-r}\)
• \(S_{\infty} = \frac{a}{1-r}\) (|r| < 1)