Arithmetic and geometric progressions are fundamental sequences used in fields such as finance, physics and computer science. In an arithmetic progression, the difference between consecutive terms is constant. In a geometric progression, each term after the first is obtained by multiplying the previous term by a fixed non-zero common ratio \(r\). Arithmetic ProgressionGeometric Progression NotationFirst term: \(a\) Common difference: \(d\) First term: \(a\) Common ratio: \(r\) nth term\[T_n=a+(n-1)d\] \[u_n=ar^{n-1}\] \[u_1=S_1,\qquad u_n=S_n-S_{n-1}\quad(n\geq2)\] Equivalently, define \(S_0=0\). Sum, \(S_n\)\[S_n=\frac{n}{2}\left[2a+(n-1)d\right]\] \[S_n=\frac{n}{2}\left[\text{First term}+\text{Last term}\right]\] \[S_n=\frac{a(r^n-1)}{r-1}\quad\text{or}\quad S_n=\frac{a(1-r^n)}{1-r},\qquad r\ne1\] When \(r=1\), \(S_n=na\). Sum to infinity, \(S_\infty\)A nonzero arithmetic series does not converge. If \(a=d=0\), every partial sum is zero, so the sum is zero.The series converges for \(\lvert r\rvert<1\). \[S_\infty=\frac{a}{1-r}\] Proving\[T_n-T_{n-1}=\text{constant}\] (common difference) \[\frac{u_n}{u_{n-1}}=\text{constant}\] (common ratio) MeanIf \(x,y,z\) are consecutive terms: \[y-x=z-y\quad\Rightarrow\quad y=\frac{x+z}{2}\] If \(x,y,z\) are consecutive terms: \[\frac{y}{x}=\frac{z}{y}\quad\Rightarrow\quad y=\pm\sqrt{xz}\] The ratio tests \(u_n/u_{n-1}\) and \(y/x=z/y\) require nonzero denominators. To include the zero sequence, use the recurrence \(u_n=ru_{n-1}\) instead. The geometric mean ratio statement assumes \(x,y,z\ne0\).

Arithmetic and Geometric Progressions

Arithmetic and geometric progressions are fundamental sequences used in fields such as finance, physics and computer science. In an arithmetic progression, the difference between consecutive terms is constant. In a geometric progression, each term after the first is obtained by multiplying the previous term by a fixed non-zero common ratio \(r\).

	Arithmetic Progression	Geometric Progression
Notation	First term: \(a\) Common difference: \(d\)	First term: \(a\) Common ratio: \(r\)
nth term	\[T_n=a+(n-1)d\]	\[u_n=ar^{n-1}\]
nth term	\[u_1=S_1,\qquad u_n=S_n-S_{n-1}\quad(n\geq2)\] Equivalently, define \(S_0=0\).
Sum, \(S_n\)	\[S_n=\frac{n}{2}\left[2a+(n-1)d\right]\] \[S_n=\frac{n}{2}\left[\text{First term}+\text{Last term}\right]\]	For \(r\ne1\): \[S_n=\frac{a(r^n-1)}{r-1}\] \[S_n=\frac{a(1-r^n)}{1-r}\] When \(r=1\), \(S_n=na\).
Sum to infinity, \(S_\infty\)	A nonzero arithmetic series does not converge. If \(a=d=0\), every partial sum is zero, so the sum is zero.	The series converges for \(\lvert r\rvert<1\). \[S_\infty=\frac{a}{1-r}\]
Proving	\[T_n-T_{n-1}=\text{constant}\] (common difference)	\[\frac{u_n}{u_{n-1}}=\text{constant}\] (common ratio)
Mean	If \(x,y,z\) are consecutive terms: \[y-x=z-y\quad\Rightarrow\quad y=\frac{x+z}{2}\]	If \(x,y,z\) are consecutive terms: \[\frac{y}{x}=\frac{z}{y}\quad\Rightarrow\quad y=\pm\sqrt{xz}\]

The ratio tests \(u_n/u_{n-1}\) and \(y/x=z/y\) require nonzero denominators. To include the zero sequence, use the recurrence \(u_n=ru_{n-1}\) instead. The geometric mean ratio statement assumes \(x,y,z\ne0\).

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