Chapter 12: Series
12.1 Arithmetic Progressions (AP)
An arithmetic progression is a sequence in which the difference between consecutive terms is constant. This constant difference is called the common difference \(d\).
General form:
\[
a, \ a+d, \ a+2d, \ a+3d, \ldots
\]
where \(a\) is the first term and \(d\) is the common difference.
Derivation of the nth term formula:
The \(n\)-th term \(u_n\) is obtained by adding \(d\) to \(a\) for each step. After \((n-1)\) steps:
\[
u_n = a + (n-1)d
\]
Derivation of the sum of the first \(n\) terms:
Write the sum forward and backward:
\[
S_n = a + (a+d) + (a+2d) + \cdots + (a+(n-1)d)
\]
\[
S_n = (a+(n-1)d) + (a+(n-2)d) + \cdots + a
\]
Adding corresponding terms gives \(2S_n = n[2a + (n-1)d]\). Thus:
\[
S_n = \frac{n}{2}[2a + (n-1)d]
\]
Alternatively, if the last term \(l = a + (n-1)d\) is known:
\[
S_n = \frac{n}{2}(a + l)
\]
AP Formulas:
\(u_n = a + (n-1)d\)
\(S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + l)\)
Worked Example 12.1 (Finding terms – non-calculator)
An AP has first term 3 and common difference 7. Find the 10th term and the sum of the first 12 terms.
Solution:
(a) \(u_{10} = 3 + (10-1) \times 7 = 3 + 63 = 66\)
(b) \(S_{12} = \frac{12}{2}[2(3) + (12-1) \times 7] = 6[6 + 77] = 6 \times 83 = 498\)
Worked Example 12.2 (Finding \(a\) and \(d\))
The 3rd term of an AP is 12 and the 7th term is 28. Find the first term and common difference.
Solution:
\(u_3 = a + 2d = 12\)
\(u_7 = a + 6d = 28\)
Subtract: \((a + 6d) - (a + 2d) = 28 - 12 \Rightarrow 4d = 16 \Rightarrow d = 4\)
Substitute: \(a + 2(4) = 12 \Rightarrow a = 4\)
Answer: \(a = 4\), \(d = 4\)
Worked Example 12.3 (Mauritian context – savings plan)
A student saves Rs 50 in the first week, Rs 70 in the second week, Rs 90 in the third week, and so on. How much will they save in the 20th week, and what is the total saved after 20 weeks?
Solution:
This is an AP with \(a = 50\), \(d = 20\).
20th week: \(u_{20} = 50 + 19 \times 20 = 50 + 380 = \text{Rs } 430\)
Total after 20 weeks: \(S_{20} = \frac{20}{2}[2(50) + 19 \times 20] = 10[100 + 380] = 10 \times 480 = \text{Rs } 4800\)
12.2 Geometric Progressions (GP)
A geometric progression is a sequence in which the ratio between consecutive terms is constant. This constant ratio is called the common ratio \(r\).
General form:
\[
a, \ ar, \ ar^2, \ ar^3, \ldots
\]
where \(a\) is the first term and \(r\) is the common ratio.
Derivation of the nth term formula:
The \(n\)-th term \(u_n\) is obtained by multiplying \(a\) by \(r\) for each step. After \((n-1)\) multiplications:
\[
u_n = ar^{n-1}
\]
Derivation of the sum of the first \(n\) terms:
Write the sum:
\[
S_n = a + ar + ar^2 + \cdots + ar^{n-1}
\]
Multiply by \(r\): \(rS_n = ar + ar^2 + \cdots + ar^n\)
Subtract: \(S_n - rS_n = a - ar^n \Rightarrow S_n(1 - r) = a(1 - r^n)\)
Thus, for \(r \neq 1\):
\[
S_n = \frac{a(1 - r^n)}{1 - r}
\]
GP Formulas:
\(u_n = ar^{n-1}\)
\(S_n = \frac{a(1 - r^n)}{1 - r} \quad (r \neq 1)\)
Worked Example 12.4 (Finding terms – non-calculator)
A GP has first term 5 and common ratio 3. Find the 4th term and the sum of the first 6 terms.
Solution:
\(u_4 = 5 \times 3^{4-1} = 5 \times 27 = 135\)
\(S_6 = \frac{5(1 - 3^6)}{1 - 3} = \frac{5(1 - 729)}{-2} = \frac{5(-728)}{-2} = 1820\)
Worked Example 12.5 (Finding \(a\) and \(r\))
The 2nd term of a GP is 12 and the 5th term is 96. Find the first term and common ratio.
Solution:
\(u_2 = ar = 12\)
\(u_5 = ar^4 = 96\)
Divide: \(\frac{ar^4}{ar} = \frac{96}{12} \Rightarrow r^3 = 8 \Rightarrow r = 2\)
Substitute: \(a(2) = 12 \Rightarrow a = 6\)
Answer: \(a = 6\), \(r = 2\)
Worked Example 12.6 (Mauritian context – population growth)
The population of a village in Mauritius doubles every year. If the population is 500 in 2024, find the population in 2030.
Solution:
This is a GP with \(a = 500\), \(r = 2\).
2030 is 6 years after 2024, so \(n = 7\) (2024 is term 1).
\(u_7 = 500 \times 2^{6} = 500 \times 64 = 32,000\)
Answer: 32,000 people.
12.3 Sum to Infinity of a Convergent GP
A GP converges (has a finite sum to infinity) if and only if \(|r| < 1\).
Derivation of the sum to infinity:
For \(|r| < 1\), as \(n \to \infty\), \(r^n \to 0\).
Thus:
\[
S_\infty = \lim_{n \to \infty} \frac{a(1 - r^n)}{1 - r} = \frac{a}{1 - r}
\]
\(S_\infty = \frac{a}{1 - r} \quad (|r| < 1)\)
Worked Example 12.7 (Sum to infinity – non-calculator)
Find the sum to infinity of the GP \(20 + 5 + 1.25 + \cdots\).
Solution:
First term \(a = 20\). Common ratio \(r = \frac{5}{20} = \frac{1}{4}\).
Since \(|r| = \frac{1}{4} < 1\), the sum to infinity exists.
\[
S_\infty = \frac{20}{1 - \frac{1}{4}} = \frac{20}{\frac{3}{4}} = 20 \times \frac{4}{3} = \frac{80}{3}
\]
Worked Example 12.8 (Finding \(a\) from \(S_\infty\))
A convergent GP has sum to infinity 24 and common ratio \(-\frac{1}{3}\). Find the first term.
Solution:
\[
S_\infty = \frac{a}{1 - r} \Rightarrow 24 = \frac{a}{1 - (-\frac{1}{3})} = \frac{a}{\frac{4}{3}} \Rightarrow a = 24 \times \frac{4}{3} = 32
\]
12.4 The Binomial Theorem for Positive Integer \(n\)
The Binomial Theorem states that for a positive integer \(n\):
\[
(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r
\]
where \(\binom{n}{r} = \frac{n!}{(n-r)!r!}\) are the binomial coefficients.
Expanded form:
\[
(a + b)^n = a^n + \binom{n}{1} a^{n-1}b + \binom{n}{2} a^{n-2}b^2 + \cdots + \binom{n}{r} a^{n-r}b^r + \cdots + b^n
\]
\((a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\)
Worked Example 12.9 (Expanding – non-calculator)
Expand \((2x + 3)^4\).
Solution:
\[
(2x + 3)^4 = \binom{4}{0}(2x)^4 + \binom{4}{1}(2x)^3(3) + \binom{4}{2}(2x)^2(3)^2 + \binom{4}{3}(2x)(3)^3 + \binom{4}{4}(3)^4
\]
\[
= 16x^4 + 4(8x^3)(3) + 6(4x^2)(9) + 4(2x)(27) + 81
\]
\[
= 16x^4 + 96x^3 + 216x^2 + 216x + 81
\]
12.5 General Term \(\binom{n}{r} a^{n-r} b^r\)
The (r+1)-th term in the expansion of \((a + b)^n\) is:
\[
T_{r+1} = \binom{n}{r} a^{n-r} b^r
\]
This is useful for finding specific terms without fully expanding.
\(T_{r+1} = \binom{n}{r} a^{n-r} b^r\)
Worked Example 12.11 (Finding a specific term – non-calculator)
Find the term containing \(x^6\) in the expansion of \((x + 2)^8\).
Solution:
For \((x + 2)^8\), \(a = x\), \(b = 2\), \(n = 8\). The term with \(x^6\) occurs when \(n - r = 6 \Rightarrow r = 2\).
\[
T_3 = \binom{8}{2} x^{6} 2^2 = 28 \times 4 \times x^6 = 112x^6
\]
Worked Example 12.12 (Finding term independent of \(x\))
Find the term independent of \(x\) in \(\left(x + \frac{1}{x}\right)^4\).
Solution:
General term: \(T_{r+1} = \binom{4}{r} x^{4-r} \left(\frac{1}{x}\right)^r = \binom{4}{r} x^{4-2r}\)
For the term independent of \(x\), \(4 - 2r = 0 \Rightarrow r = 2\).
\[
T_3 = \binom{4}{2} x^{0} = 6
\]
Answer: 6.
Chapter 12 Summary
| Concept |
Key points |
| AP nth term |
\(u_n = a + (n-1)d\) |
| AP sum |
\(S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + l)\) |
| GP nth term |
\(u_n = ar^{n-1}\) |
| GP sum |
\(S_n = \frac{a(1 - r^n)}{1 - r}\) (for \(r \neq 1\)) |
| Sum to infinity |
\(S_\infty = \frac{a}{1 - r}\) (only if \(|r| < 1\)) |
| Binomial expansion |
\((a + b)^n = \sum \binom{n}{r} a^{n-r} b^r\) |
| General term |
\(T_{r+1} = \binom{n}{r} a^{n-r} b^r\) |
Exercises – Chapter 12
Easy (Drill – Non-Calculator)
- In an AP, \(a = 5\), \(d = 3\). Find:
(a) The 8th term
(b) The sum of the first 10 terms
- In a GP, \(a = 4\), \(r = 2\). Find:
(a) The 5th term
(b) The sum of the first 4 terms
- Expand \((x + 2)^3\).
Medium (Examination Style)
- The 4th term of an AP is 22 and the 10th term is 46. Find \(a\) and \(d\).
- A GP has first term 3 and common ratio 2. Find the sum of the first 8 terms.
- Find the sum to infinity of the GP \(16 + 8 + 4 + \cdots\).
- Find the coefficient of \(x^3\) in the expansion of \((1 + 2x)^5\).
- Mauritian context: The number of tourists visiting a hotel increases by 10% each year. If there were 500 tourists in 2023, how many tourists will there be in 2028?
Hard (Challenge for A*)
- The 2nd and 5th terms of a GP are 6 and 162 respectively. Find:
(a) The common ratio
(b) The first term
(c) The sum of the first 6 terms
- Find the term independent of \(x\) in the expansion of \(\left(x^2 + \frac{1}{x}\right)^6\).
- In the expansion of \((1 + kx)^7\), the coefficient of \(x^3\) is 280. Find the value of \(k\).