Topic 11 Arithmetic and geometric progressions

11.1 Arithmetic progressions

An arithmetic progression (AP) is a (finite or infinite) sequence of numbers \[ a_1,a_2,a_3,\dots \] such that the difference between consecutive terms is a constant (also called the common difference) \(d\). In other words, \[ d=a_2-a_1=a_3-a_2=a_4-a_3=\dots. \] For example:

• \(1,3,5,7,\dots\) (\(d=2\))
• \(3,1,-1,-3,-5,\dots\) (\(d=-2\))

We have \[\begin{align*} a_2&=a_1+d\\ a_3&=a_2+d=a_1+2d \\ a_4&=a_3+d=a_1+3d \end{align*}\] In general: \(a_n = a+(n-1)d\) where \(a=a_1\).

Theorem 11.1 The sum \(S_n\) of the first \(n\) terms of an AP with the first term \(a\) and the common difference \(d\) is given by \[ \boxed{S_n=\frac{n}2\left(2a+(n-1)d\right)} \]

Proof (non-examinable): \[ \begin{array}{ccccccccc} S_n &=& a & + & (a+d) & + & \dots & + & (a+(n-1)d) \\ S_n &=& (a+(n-1)d) & + & (a+(n-2)d) & + & \dots & + & a\\ \implies \\ 2S_n & = & (2a+(n-1)d) & + & (2a+(n-1)d) & + & \dots & + & (2a+(n-1)d)\\ & = & n(2a+(n-1)d) \end{array} \] The “\(\implies\)” step is by summing the two previous equations “vertically”. The last step holds as there are \(n\) summands, each of them equal to \(2a+(n-1)d\)).

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

Example 11.1 Determine the sum \(S_{20}\) of the first \(20\) terms of the arithmetic progression: \(10, 6, 2, -2, \dots\).

Solution. \(a=10\), \(d=-4\), \(n=20\)
\(\implies\) \(S_{20}=\frac{20}{2}(2\cdot 10+ 19\cdot(-4))\) \(=10(20-76)=-560\).

Example 11.2 The sum \(S_8\) of the first \(8\) terms of an AP is \(90\), and its first term is \(6\). What is the common difference?

Solution. \(90=S_8=\frac{8}{2}(2\cdot 6 + 7\cdot d)\)
\(\implies\) \(90=4(12+7d)\)
\(\implies\) \(\frac{90}{4}-12=7d\)
\(\implies\) \(d=\frac{42}{28}=\frac32=1.5\).

Example 11.3 How many terms of the AP \(3,6,9,\dots\) must be taken so that their sum is \(135\)?

Solution. \(a=3\), \(d=3\), \(S_n=135\), \(n=?\)
\(\implies\) \(135 = S_n = \frac{n}{2}(2\cdot 3+(n-1)\cdot 3)\)
\(\implies\) \(270 = 3n^2 + 3n\)
\(\implies\) \(n^2+n-90 = 0\)
\(\implies\) \((n+10)(n-9)=0\)
\(\implies\) \(n=9\) (since \(n\) must be positive).

11.2 Geometric progressions

A geometric progression (GP) is a (finite or infinite) sequence of numbers \[ a_1,a_2,a_3,\dots \] such that the quotient of the consecutive terms is a constant (also called the common ratio) \(r\). In other words, \[ r = \frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \dots \] For example:

• \(1,\frac12,\frac14,\frac18,\dots\) (\(r=\frac12\))
• \(2,-6,18,-54,\dots\) (\(r=-3\))

We have: \(a_2=a_1r\),   \(a_3=a_2r = a_1r^2\),   \(a_4=a_3r=a_1r^3\),

In general: \(a_n=a\cdot r^{n-1}\) where \(a=a_1\).

Theorem 11.2 The sum \(S_n\) of the first \(n\) terms of a GP with the first term \(a\) and the common ratio \(r\) is given by \[ \boxed{S_n = a\cdot \frac{1-r^n}{1-r}} \] In particular, if \(-1<r<1\), then the sum \(S_\infty\) of all (infinitely many) terms is \(S_\infty = a\cdot\dfrac{1}{1-r}\). (The sum of an infinite GP is also called geometric series.)

Proof (non–examinable): \[\begin{align*} (1-r)S_n &= (1-r)(a+ar+\dots+ar^{n-1})\\ &= (a+ar+\dots+ar^{n-1}) - (ar + ar^2 +\dots +ar^n) \\ &= a-ar^n = a(1-r^n) \qquad\text{(everything else cancels)} \end{align*}\] Thus \(S_n = a\frac{1-r^n}{1-r}\).
Now if \(-1<r<1\), then \(\lim_{n\to\infty}r^n=0\). So \(S_\infty=\lim_{n\to\infty}S_n = a\frac{1}{1-r}\).

Example 11.4 Determine the sum \(S_7\) of the first \(7\) terms of the geometric progession \(4,-8,16,\dots\)

Solution. \(a=4\), \(r=-2\), \(n=7\)
\(\implies\) \(S_7=4\dfrac{1-(-2)^7}{1-(-2)}=4\cdot\dfrac{129}3 = 172\).

Example 11.5 A GP is given by \(\dfrac14,\dfrac1{16},\dfrac1{64},\dots\). Determine \(S_\infty\).

Solution. \(a=\frac14\), \(r=\frac14\)
\(\implies\) \(S_\infty = \dfrac{1}{4}\cdot\dfrac{1}{1-\frac14}= \dfrac{1}{4}\cdot\dfrac{1}{\frac{3}{4}}=\dfrac{1}{3}\).