An arithmetic progression or AP is any sequence where each new term is obtained by adding a constant number to the preceding term. This common number is referred to as the common difference. For example, $10, 20, 30, 40$, is an arithmetic progression increasing by $10$, or $-4, -3, -2, -1$ is an arithmetic progression decreasing by $1$ with each term.
An arithmetic progression can be expressed as $a, a+d, a+2d, a+3d, \dotso$, where $a$ denotes the first term and $d$ denotes the common difference between each term.
The $n^{\text{th} }$ term is given by \[u_n=a+(n-1)d\]
The sum of the terms of an arithmetic progression gives an arithmetic series. \[S_N = \sum_{n=1}^{N} a+(n-1)d\] A quicker formula for working out the sum of an arithmetic progression is \[S_N = \frac{1}{2}N\bigl(2a+(N-1)d\bigr)\] If the value of the last term, denoted $l$, is known instead of the common difference $d$, the series can be given by \[S_N = \frac{1}{2}N(a+l)\]
Write down the first five terms of the arithmetic progression with first term $8$ and common difference $7$.
We have $a=8$ and $d=7$.
The form of an arithmetic progression is $a, a+d, a+2d, a+3d, a+4d$, so using these values of $a$ and $d$ the first five terms are:
\[8, \; 8 + 7, \; 8 + 2\times 7, \; 8+3 \times 7, \; 8 + 4\times 7\]
\[8, 15, 22, 29, 36\]
Find the $17^{\text{th} }$ term of the arithmetic progression with first term $5$ and common difference $2$.
We have $a=5$, $d=2$ and $n=17$.
The nth term is given by $u_n = a+(n-1)d$, so
\begin{align} u_{17} &= a+ (17-1)d\\ &= 5 + 16\times 2\\ &= 37 \end{align}
Find the sum of the first fifty terms of the sequence: $1, 3, 5, 7, 9, \dotso$.
We have $a=1$, $d=2$ and $n=50$, and are asked to calculate \[\sum^{50}_{n=1} 1 + 2(n-1)\]
Using the formula $S_n=\dfrac{1}{2}n\bigl(2a+(n-1)d\bigr)$:
\begin{align} S_{50} &= \frac{1}{2}\times 50(2\times 1 + (50-1))\\ &=25(2+49)\\ &=25(2+98)\\ &=1275 \end{align}
A geometric progression, or GP, is a sequence where each new term is obtained by multiplying the preceding term by a constant number.
A geometric progression can be expressed as $a, ar, ar^2, ar^3, \dotso$, where $a$ denotes the first term and $r$ is the common ratio between terms. The $n^{\text{th} }$ term is given by \[u_n = ar^{n-1}\]
The sum of the terms of a geometric progression gives a geometric series, \[S_N = \sum^N_{n=1} ar^{n-1}.\]
A quicker formula for working out the sum of a geometric progression is \[S_N=\frac{a(1-r^N)}{1-r}\]
provided that $r \neq 1$.
A series converges if, as $N$ increases, the partial series sum $S_N$ approaches a single number.
If a sequence converges, then the infinite series sum $S_\infty$ can be calculated explicitly.
A geometric series converges if its common ratio is in the range $(-1,1)$. The infinite series sum of a convergent geometric sequence with first term $a$ and common ratio $r$ is
\[S_\infty = \frac{a}{1-r}\]
Write down the first five terms of the geometric progression which has first term $1$ and common ratio $\dfrac{1}{2}$.
We have $a=1$ and $r=\dfrac{1}{2}$.
The general form of a geometric progression is $a, ar, ar^2, ar^3, ar^4, \dotso$, so using these values of $a$ and $r$ the first five terms are:
\[1, \; 1\times \dfrac{1}{2}, \; 1\times \left(\dfrac{1}{2}\right)^2, \; 1\times \left(\dfrac{1}{2}\right)^3, \; 1\times \left(\dfrac{1}{2}\right)^4\]
\[1, \dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{8},\dfrac{1}{16}\]
Find the $10^{\text{th} }$ term of the geometric progression with first term $3$ and common ratio $2$.
We have $a=3$, $r=2$ and $n=10$.
The formula for the $n^{\text{th} }$ term is $u_n = ar^{n-1}$, so
\begin{align} u_{10} &= ar^{10-1}\\ &=3\times 2^9\\ &= 1536 \end{align}
How many terms are in the geometric progression $2, 4, 8, \dotso, 128$?
We have $a=2$ and $r=2$. We also know that the $10^{\text{th} }$ term is $128$.
Using the formula for the $n^{\text{th} }$ term $u_n = ar^{n-1}$
\begin{align} 128 &= 2\times 2^{n-1} \\ 64 &= 2^{n-1} \\ 2^6 &= 2^{n-1} \\ 6&= n-1 \\ n&=7 \end{align}
An alternative method would be to use logarithms.
\begin{align} 128 &= 2\times 2^{n-1}\\ 64 &= 2^{n-1} \\ \log 64 &= \log(2^{n-1})\\ \log 64 &= (n-1)\log 2\\ \frac{\log 64}{\log 2}&= n-1\\ 6&=n-1\\ n&=7 \end{align} So there are $7$ terms in the given geometric progression.
Find the sum of the first five terms of the geometric progression $8, -4, 2, \dotso$.
We have $a=8$, $r=-\dfrac{1}{2}$ and $N=5$. We are asked to calculate \[\sum^5_{n=1} 8 \times \left(-\frac{1}{2}\right)^{n-1}.\]
The formula for the sum of a geometric progression is $S_N = \dfrac{a(1-r^N)}{1-r}$. We want to find $S_5$.
\begin{align} S_5 &= \frac{8\Bigl(1-\left(-\frac{1}{2}\right)^5\Bigr)}{1-\left(-\frac{1}{2}\right)}\\ &=\frac{8\Bigl(1-\left(-\frac{1}{32}\right)\Bigr)}{ \frac{3}{2} }\\ &=\frac{ 2 \times 8 \times \frac{33}{32} }{3}\\ &=\frac{11}{2} \end{align}
Find the sum to infinity of the geometric progression $1, \dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27},\dotso$.
We have $a=1$ and $r=\dfrac{1}{3}$.
As $-1\lt r\lt 1$, the series converges and we can use the formula $S_\infty = \dfrac{a}{1-r}$.
\begin{align} S_\infty &= \frac{1}{ 1-\frac{1}{3} }\\ &=\frac{1}{\; \frac{2}{3}\;}\\ &=\frac{3}{2} \end{align}
Hayley Bishop finds the fifteenth term and the sum of the first twenty terms of the arithmetic progression $2,6,10,14,\dotso$.
Hayley Bishop finds the ninth term and the sum of the first five terms of the geometric progression $3,1,\frac{1}{3},\frac{1}{9}\dotso$.
Hayley Bishop finds the sum to infinity of the geometric progression $3,1,\frac{1}{3},\frac{1}{9}\dotso$.
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Test yourself: Arithmetic sequences
Test yourself: Geometric sequences