CIE AS/A Level Maths-1.6 Series- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-1.6 Series- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-1.6 Series- Study Notes
Key Concepts:
- Binomial Expansion
- Arithmetic Progressions (AP)
- Geometric Progressions (GP)
- Convergence of a Geometric Progression
Binomial Expansion
Binomial Expansion
The expansion of \((a + b)^n\), where \(n\) is a positive integer, can be written using the Binomial Theorem:
\((a + b)^n = \displaystyle \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\)
Here, the notation:
- \(\binom{n}{r}\) (read as “n choose r”) is the binomial coefficient.
- \(\binom{n}{r} = \dfrac{n!}{r!(n-r)!}\).
- \(n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1\).
Important Notes:
- The expansion has \(n+1\) terms.
- The powers of \(a\) decrease from \(n\) down to 0.
- The powers of \(b\) increase from 0 up to \(n\).
Example:
Expand \((a+b)^4\).
▶️ Answer/Explanation
Step 1: Write the general expansion
\((a+b)^4 = \sum_{r=0}^{4} \binom{4}{r} a^{4-r} b^r\).
Step 2: Compute each term
- \(r=0: \binom{4}{0} a^4 = 1 \cdot a^4 = a^4\)
- \(r=1: \binom{4}{1} a^3 b = 4a^3b\)
- \(r=2: \binom{4}{2} a^2 b^2 = 6a^2 b^2\)
- \(r=3: \binom{4}{3} ab^3 = 4ab^3\)
- \(r=4: \binom{4}{4} b^4 = b^4\)
Final Expansion:
\((a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\).
Example:
Find the coefficient of \(x^3\) in the expansion of \((2+x)^5\).
▶️ Answer/Explanation
Step 1: General term
\(T_{r+1} = \binom{5}{r} (2)^{5-r} x^r\).
Step 2: We need the term with \(x^3\)
This means \(r = 3\).
\(T_4 = \binom{5}{3} (2)^{2} x^3\).
Step 3: Simplify
\(\binom{5}{3} = 10\), so \(T_4 = 10 \cdot 4 \cdot x^3 = 40x^3\).
Answer: Coefficient of \(x^3\) is \(40\).
Arithmetic Progressions (AP)
Arithmetic Progressions (AP)
An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant is called the common difference \(d\).
General form of an AP:
\(a, a+d, a+2d, a+3d, \dots\)
- \(a\) = the first term
- \(d\) = the common difference
- \(n\) = number of terms
Important Formulas:
- nth term: \(u_n = a + (n-1)d\)
- Sum of first \(n\) terms: \(S_n = \dfrac{n}{2} \big[2a + (n-1)d \big]\)
- Alternative form: \(S_n = \dfrac{n}{2}(a + u_n)\)
Three Numbers in AP:
If three numbers are in AP, they can be written as:
\((a-d), \; a, \; (a+d)\)
Another useful property: If \(a, b, c\) are in AP, then
\(2b = a + c\).
This property is frequently used in solving algebraic and number problems.
Example:
Find the 15th term of the AP: \(7, 10, 13, 16, \dots\).
▶️ Answer/Explanation
Step 1: Identify terms
\(a = 7\), \(d = 3\).
Step 2: Use formula for nth term
\(u_{15} = a + (15-1)d = 7 + 14 \times 3\).
\(u_{15} = 7 + 42 = 49\).
Answer: The 15th term is \(49\).
Example:
The 5th term of an AP is 18 and the 12th term is 39. Find the first term and the common difference.
▶️ Answer/Explanation
Step 1: Use nth term formula
\(u_n = a + (n-1)d\).
For 5th term: \(a + 4d = 18\). (Equation 1)
For 12th term: \(a + 11d = 39\). (Equation 2)
Step 2: Subtract equations
\((a+11d) – (a+4d) = 39 – 18\).
\(7d = 21 \implies d = 3\).
Step 3: Substitute into Equation 1
\(a + 4(3) = 18 \implies a = 6\).
Answer: First term \(a = 6\), common difference \(d = 3\).
Example:
Find the sum of the first 20 terms of the AP: \(5, 8, 11, \dots\).
▶️ Answer/Explanation
Step 1: Identify terms
\(a = 5\), \(d = 3\), \(n = 20\).
Step 2: Use sum formula
\(S_n = \dfrac{n}{2}[2a + (n-1)d]\).
\(S_{20} = \dfrac{20}{2}[2(5) + 19 \cdot 3]\).
\(S_{20} = 10[10 + 57] = 10 \cdot 67 = 670\).
Answer: The sum of the first 20 terms is \(670\).
Example:
If three numbers are in AP such that their sum is 27 and the middle term is 9, find the numbers.
▶️ Answer/Explanation
Step 1: Use the property
Let the three numbers be \(a, b, c\) in AP.
Then \(2b = a + c\).
Step 2: Use given conditions
Sum: \(a + b + c = 27\).
Middle term: \(b = 9\).
Step 3: Substitute
\(a + 9 + c = 27 \implies a + c = 18\).
But \(2b = a + c \implies 2(9) = a + c \implies a + c = 18\).
This is consistent. Possible numbers: \((a, b, c) = (6, 9, 12)\).
Answer: The numbers are \(6, 9, 12\) (in AP).
Geometric Progressions (GP)
Geometric Progressions (GP)
A Geometric Progression (GP) is a sequence of numbers in which each term is obtained by multiplying the previous term by a constant. This constant is called the common ratio \(r\).
General form of a GP:
\(a, ar, ar^2, ar^3, \dots\)
- \(a\) = the first term
- \(r\) = the common ratio
- \(n\) = number of terms
Important Formulas:
- nth term: \(u_n = ar^{\,n-1}\)
- Sum of first \(n\) terms (when \(r \neq 1\)): \(S_n = \dfrac{a(r^n – 1)}{r – 1}\)
Three Numbers in GP:
If three numbers are in GP, they can be written as:
\(\dfrac{a}{r}, \; a, \; ar\)
Another useful property: If \(a, b, c\) are in GP, then
\(b^2 = ac\).
This property is very useful in solving algebraic and number problems.
Example:
Find the 6th term of the GP: \(3, 6, 12, 24, \dots\)
▶️ Answer/Explanation
Step 1: Identify values
\(a = 3, \; r = \dfrac{6}{3} = 2\).
Step 2: Apply formula
\(u_6 = ar^{\,6-1} = 3 \times 2^5 = 3 \times 32 = 96\).
Answer: The 6th term is \(96\).
Example:
Find the sum of the first 5 terms of the GP: \(2, 6, 18, 54, \dots\)
▶️ Answer/Explanation
Step 1: Identify values
\(a = 2, \; r = 3, \; n = 5\).
Step 2: Apply sum formula
\(S_5 = \dfrac{a(r^5 – 1)}{r – 1} = \dfrac{2(3^5 – 1)}{3 – 1}\).
\(= \dfrac{2(243 – 1)}{2} = \dfrac{2 \times 242}{2} = 242\).
Answer: The sum is \(242\).
Example:
If three numbers are in GP and their product is \(216\) while the middle term is \(6\), find the numbers.
▶️ Answer/Explanation
Step 1: Use the property
Let the three numbers be \(a, b, c\) in GP.
Then \(b^2 = ac\).
Step 2: Use given conditions
Product: \(abc = 216\).
Middle term: \(b = 6\).
Step 3: Substitute
Then \(a \cdot 6 \cdot c = 216 \implies ac = 36\).
Also, from property: \(b^2 = ac \implies 36 = ac\).
So both conditions are satisfied.
Possible numbers: \((a, b, c) = (2, 6, 18)\).
Answer: The numbers are \(2, 6, 18\) (in GP).
Example:
The first three terms of an Arithmetic Progression (AP) are \(2, 5, 8\). The first three terms of a Geometric Progression (GP) are \(3, 6, 12\).
(a) Find the 10th term of the AP.
(b) Find the sum of the first 5 terms of the GP.
(c) Compare the values from (a) and (b).
▶️ Answer/Explanation
Part (a): 10th term of the AP
AP: \(2, 5, 8, \dots\)
First term \(a = 2\), common difference \(d = 3\).
Formula: \(u_n = a + (n-1)d\).
\(u_{10} = 2 + (10-1)\cdot 3 = 2 + 27 = 29\).
Answer (a): 10th term = \(29\).
Part (b): Sum of the first 5 terms of the GP
GP: \(3, 6, 12, \dots\)
First term \(a = 3\), common ratio \(r = 2\), number of terms \(n = 5\).
Formula: \(S_n = \dfrac{a(r^n – 1)}{r – 1}\).
\(S_5 = \dfrac{3(2^5 – 1)}{2 – 1} = 3(32 – 1) = 3 \times 31 = 93\).
Answer (b): Sum = \(93\).
Part (c): Comparison
10th term of AP = \(29\).
Sum of first 5 terms of GP = \(93\).
Clearly, the GP sum grows faster than the AP term.
Final Answer: AP 10th term = \(29\), GP sum = \(93\).
Convergence of a Geometric Progression
Convergence of a Geometric Progression
A geometric progression (GP) has terms:
\(a, ar, ar^2, ar^3, \dots\)
where \(a\) is the first term and \(r\) is the common ratio.
Condition for Convergence:
A GP converges (has a finite sum to infinity) if and only if:
\(|r| < 1\).
If \(|r| \geq 1\), the series does not converge (it grows without bound or oscillates).
Formula for the Sum to Infinity:
If \(|r| < 1\), then the sum to infinity is:
\(S_\infty = \dfrac{a}{1 – r}\).
Example:
For the GP \(5, 2.5, 1.25, \dots\):
(a) Check if the GP converges.
(b) Find the sum to infinity if it converges.
▶️ Answer/Explanation
Step 1: Identify \(a\) and \(r\)
\(a = 5\), \(r = \dfrac{2.5}{5} = 0.5\).
Step 2: Check convergence condition
\(|r| = 0.5 < 1\), so the GP converges.
Step 3: Apply formula
\(S_\infty = \dfrac{a}{1 – r} = \dfrac{5}{1 – 0.5} = \dfrac{5}{0.5} = 10\).
Final Answer: The GP converges, and \(S_\infty = 10\).
Example:
Does the GP \(3, -6, 12, -24, \dots\) converge? If yes, find \(S_\infty\).
▶️ Answer/Explanation
Step 1: Identify \(a\) and \(r\)
\(a = 3\), \(r = \dfrac{-6}{3} = -2\).
Step 2: Check convergence
\(|r| = 2 \geq 1\). The GP does not converge.
Final Answer: The GP diverges, so there is no finite sum to infinity.