Arithmetic Sequences & Series
Common difference · nth term · Sum of n terms · §11.1
Key formulas
nth Term
\(u_n = a + (n-1)d\)
a = first term, d = common difference, n = term number
Sum of n Terms
\(S_n = \dfrac{n}{2}(2a+(n-1)d)\)
Or equivalently: \(S_n = \dfrac{n}{2}(a+l)\) where l = last term
Common Difference
\(d = u_{n+1} - u_n\)
Constant difference between consecutive terms. Can be positive (increasing) or negative (decreasing).
AP: first term 3, common difference 5. Find u\(_8\) and S\(_8\).
nth term
\(u_8 = 3 + (8-1)(5) = 3 + 35 = 38\)
Sum
\(S_8 = \dfrac{8}{2}(2(3)+(8-1)(5)) = 4(6+35) = 4 \times 41\)
✓
\(u_8=38,\quad S_8=164\)
Finding a and d from two terms
If given \(u_p\) and \(u_q\), set up two equations using \(u_n = a+(n-1)d\) and solve simultaneously.
\(u_3=11\) and \(u_7=27\). Find a and d.
Eq â‘
\(a+2d=11\)
Eq â‘¡
\(a+6d=27\)
②−â‘
\(4d=16 \Rightarrow d=4,\; a=3\)
Geometric Sequences & Series
Common ratio · nth term · Sum formulas · §11.2
Key formulas
nth Term
\(u_n = ar^{n-1}\)
a = first term, r = common ratio (multiply by r each time)
Sum of n Terms
\(S_n = \dfrac{a(1-r^n)}{1-r}\quad(r\neq1)\)
Or \(\dfrac{a(r^n-1)}{r-1}\). Use whichever avoids negative denominators.
Common Ratio
\(r = \dfrac{u_{n+1}}{u_n}\)
Constant ratio of consecutive terms. If |r| < 1, terms shrink. If |r| > 1, terms grow.
GP: a=2, r=3. Find \(u_5\) and \(S_5\).
nth term
\(u_5 = 2 \times 3^4 = 2 \times 81 = 162\)
Sum
\(S_5 = \dfrac{2(3^5-1)}{3-1} = \dfrac{2(243-1)}{2} = 242\)
✓
\(u_5=162,\quad S_5=242\)
Sum to infinity (|r| < 1)
\(S_\infty = \dfrac{a}{1-r}\quad(|r|<1)\)
Only exists when \(|r| < 1\) — the series converges. If \(|r| \geq 1\), the sum to infinity is undefined.
a=8, r=0.5. Find \(S_\infty\).
Check
|0.5| < 1 ✓ — converges
✓
\(S_\infty = \dfrac{8}{1-0.5} = 16\)
The Binomial Theorem
Pascal's triangle · General term · Expanding (a+b)^n · §11.3
The binomial expansion
\((a+b)^n = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r\)
Each term has form \(\binom{n}{r}a^{n-r}b^r\) where r goes from 0 to n.
General term (r+1)th term: \(T_{r+1} = \binom{n}{r}a^{n-r}b^r\)
Powers check: In each term, the powers of a and b must sum to n. If they don't — you have an error.
Expand \((2+3x)^4\) fully
Row 4
\(\binom{4}{0},\binom{4}{1},\binom{4}{2},\binom{4}{3},\binom{4}{4}\) = 1, 4, 6, 4, 1
Expand
\(1(2)^4 + 4(2)^3(3x) + 6(2)^2(3x)^2 + 4(2)(3x)^3 + 1(3x)^4\)
Simplify
\(16 + 96x + 216x^2 + 216x^3 + 81x^4\)
✓
\(16+96x+216x^2+216x^3+81x^4\)
Find the coefficient of \(x^3\) in \((1+2x)^6\)
Identify r
\(b^r = (2x)^3\) so r = 3
General term
\(T_4 = \binom{6}{3}(1)^3(2x)^3 = 20 \times 8x^3\)
✓
Coefficient = 160
Practice Questions
3 Easy · 3 Medium · 3 Hard — AP & GP
Easy 1
[2] An AP has first term 5 and common difference 3. Find the 20th term.
SOLUTION
Formula
\(a_n=a+(n-1)d=5+19(3)=\boldsymbol{62}\)
Easy 2
[2] A GP has first term 4 and common ratio 3. Find the 5th term.
SOLUTION
Formula
\(a_n=ar^{n-1}=4\times3^4=4\times81=\boldsymbol{324}\)
Easy 3
[2] Find the sum of the first 10 terms of the AP: 2, 5, 8, ...
SOLUTION
Formula
\(S_n=\tfrac{n}{2}(2a+(n-1)d)=\tfrac{10}{2}(4+27)=5\times31=\boldsymbol{155}\)
Medium 1
[4] The 4th term of a GP is 54 and the 7th term is 1458. Find the first term and common ratio.
SOLUTION
Ratio
\(\dfrac{ar^6}{ar^3}=r^3=\dfrac{1458}{54}=27\Rightarrow r=3\)
First term
\(ar^3=54\Rightarrow a(27)=54\Rightarrow a=\boldsymbol{2}\)
Medium 2
[4] Find the sum to infinity of the GP with first term 12 and common ratio \(\frac{1}{3}\).
SOLUTION
Condition
\(|r|=\tfrac13<1\) so series converges.
Formula
\(S_\infty=\dfrac{a}{1-r}=\dfrac{12}{1-\frac13}=\dfrac{12}{\frac23}=\boldsymbol{18}\)
Medium 3
[4] An AP has 20 terms. The first term is 8 and the last term is 65. Find the common difference and the sum.
SOLUTION
Find d
\(a_{20}=8+19d=65\Rightarrow d=\boldsymbol{3}\)
Sum
\(S_{20}=\tfrac{20}{2}(8+65)=10\times73=\boldsymbol{730}\)
Hard 1
[5] The sum of the first \(n\) terms of a series is \(S_n=3n^2+n\). Find the \(n\)th term and show the series is arithmetic.
SOLUTION
nth term
\(a_n=S_n-S_{n-1}=(3n^2+n)-(3(n-1)^2+(n-1))\)
\(=3n^2+n-3n^2+6n-3-n+1=6n-2\)
\(=3n^2+n-3n^2+6n-3-n+1=6n-2\)
Arithmetic?
\(a_n=6n-2\) is linear in \(n\Rightarrow\) constant difference \(d=6\). It is an AP with \(d=\boldsymbol{6}\).
Hard 2
[5] The 2nd, 5th, and 14th terms of an AP are consecutive terms of a GP. Given the first term of the AP is 2, find the common difference.
SOLUTION
AP terms
\(a_2=2+d,\; a_5=2+4d,\; a_{14}=2+13d\)
GP condition
Ratio constant: \(\dfrac{2+4d}{2+d}=\dfrac{2+13d}{2+4d}\)
\((2+4d)^2=(2+d)(2+13d)\)
\(4+16d+16d^2=4+28d+2d+13d^2\)
\(3d^2-14d=0\Rightarrow d(3d-14)=0\Rightarrow d=\dfrac{14}{3}\) (reject \(d=0\))
\((2+4d)^2=(2+d)(2+13d)\)
\(4+16d+16d^2=4+28d+2d+13d^2\)
\(3d^2-14d=0\Rightarrow d(3d-14)=0\Rightarrow d=\dfrac{14}{3}\) (reject \(d=0\))
Hard 3
[5] Find the least value of \(n\) for which \(\displaystyle\sum_{k=1}^n 3^k > 10^6\).
SOLUTION
GP sum
\(S_n=\dfrac{3(3^n-1)}{3-1}=\dfrac{3^{n+1}-3}{2}>10^6\)
Solve
\(3^{n+1}>2\times10^6+3\approx2000003\)
\((n+1)\ln3>\ln2000003\)
\(n+1>12.57\Rightarrow n>11.57\Rightarrow\boldsymbol{n=12}\)
\((n+1)\ln3>\ln2000003\)
\(n+1>12.57\Rightarrow n>11.57\Rightarrow\boldsymbol{n=12}\)
Past Year Paper Questions
Cambridge 0606 — AP & GP
0606
[10]
(a) The sum of the first \(n\) terms of an AP is \(2n^2+3n\). Find the first term and common difference. [4]
(b) A GP has sum to infinity 60 and sum of first 3 terms 52. Find the common ratio. [4]
(c) Hence state whether the GP is convergent and justify. [2]
(b) A GP has sum to infinity 60 and sum of first 3 terms 52. Find the common ratio. [4]
(c) Hence state whether the GP is convergent and justify. [2]
FULL SOLUTION
(a) nth term
\(a_n=S_n-S_{n-1}=(2n^2+3n)-(2(n-1)^2+3(n-1))=4n+1\)
First term (\(n=1\)): \(a_1=5\). Common difference: \(d=4\).
First term (\(n=1\)): \(a_1=5\). Common difference: \(d=4\).
(b) Setup
\(S_\infty=\dfrac{a}{1-r}=60\) ...[1] \(S_3=a\cdot\dfrac{1-r^3}{1-r}=52\) ...[2]
[2]÷[1]: \(1-r^3=\dfrac{52}{60}=\dfrac{13}{15}\Rightarrow r^3=\dfrac{2}{15}\)
[2]÷[1]: \(1-r^3=\dfrac{52}{60}=\dfrac{13}{15}\Rightarrow r^3=\dfrac{2}{15}\)
(b) Solve
\(r=\sqrt[3]{\dfrac{2}{15}}\approx0.51\)
(c)
\(|r|\approx0.51<1\Rightarrow\) series converges. ✓
0606 Style
Binomial — Specific Term
[8]
(a) Find the coefficient of \(x^3\) in the expansion of \((3-2x)^6\). [3]
(b) Find the term independent of \(x\) in \(\left(x^2-\dfrac{2}{x}\right)^9\). [4]
(c) If the coefficient of \(x^2\) in \((1+kx)^5\) is 40, find \(k\). [1]
(b) Find the term independent of \(x\) in \(\left(x^2-\dfrac{2}{x}\right)^9\). [4]
(c) If the coefficient of \(x^2\) in \((1+kx)^5\) is 40, find \(k\). [1]
FULL WORKED SOLUTION
(a)
\(T_{r+1}=\binom{6}{r}3^{6-r}(-2x)^r\). Need \(r=3\):
\(\binom{6}{3}3^3(-2)^3=20\times27\times(-8)=\boldsymbol{-4320}\)
\(\binom{6}{3}3^3(-2)^3=20\times27\times(-8)=\boldsymbol{-4320}\)
(b)
\(T_{r+1}=\binom{9}{r}(x^2)^{9-r}\left(-\frac{2}{x}\right)^r=\binom{9}{r}(-2)^rx^{18-2r-r}\)
Need \(18-3r=0\Rightarrow r=6\).
\(\binom{9}{6}(-2)^6=84\times64=\boldsymbol{5376}\)
Need \(18-3r=0\Rightarrow r=6\).
\(\binom{9}{6}(-2)^6=84\times64=\boldsymbol{5376}\)
(c)
\(\binom{5}{2}k^2=10k^2=40\Rightarrow k^2=4\Rightarrow\boldsymbol{k=\pm2}\)
0606 Style
AP/GP Combined Problem
[7]
The 3rd term of an AP is 16 and the 8th term is 36. The 3rd, 4th and 7th terms of this AP form a geometric sequence.
(a) Find the first term and common difference of the AP. [3]
(b) Verify that the 3rd, 4th, and 7th terms form a GP and find the common ratio. [4]
(a) Find the first term and common difference of the AP. [3]
(b) Verify that the 3rd, 4th, and 7th terms form a GP and find the common ratio. [4]
FULL WORKED SOLUTION
(a)
\(a+2d=16\) and \(a+7d=36\). Subtract: \(5d=20\Rightarrow d=4\). \(a=16-8=8\).
\(\boldsymbol{a=8, d=4}\)
\(\boldsymbol{a=8, d=4}\)
(b)
3rd term: \(T_3=16\). 4th: \(T_4=20\). 7th: \(T_7=32\).
GP check: \(\frac{20}{16}=\frac{5}{4}\) and \(\frac{32}{20}=\frac{8}{5}\). These are not equal, so these do NOT form a GP. (Re-examine the problem's wording — use \(T_3, T_5, T_9\) for a typical exam version where the ratio works out.)
GP check: \(\frac{20}{16}=\frac{5}{4}\) and \(\frac{32}{20}=\frac{8}{5}\). These are not equal, so these do NOT form a GP. (Re-examine the problem's wording — use \(T_3, T_5, T_9\) for a typical exam version where the ratio works out.)