**Definition of an Arithmetic Progression (AP)**

An arithmetic progression is a sequence of numbers (infinite or finite) where, starting from the second term, each term is the sum of the preceding term and a constant \(d\). In other words:

\({u_{n + 1}} = {u_n} + d{\rm{\;}}\left( {n \in {\mathbb{N}^{\rm{*}}}} \right){\rm{.\;}}\)

The number \(d\) is called the common difference of the arithmetic progression.

**Observation:** If \(\left( {{u_n}} \right)\) is an arithmetic progression, then, from the second term onwards (excluding the last term for a finite AP), each term is the average of the two terms adjacent to it in the sequence. In other words:

\({u_k} = \dfrac{{{u_{k – 1}} + {u_{k + 1}}}}{2}\left( {k \ge 2} \right)\).

**General Term of an Arithmetic Progression**

*Theorem 1*

If an arithmetic progression \(\left( {{u_n}} \right)\) has the first term \({u_1}\) and a common difference \(d\), then the general term \({u_n}\) is determined by the formula: \({u_n} = {u_1} + \left( {n – 1} \right)d, n \ge 2.\)

**Sum of the First n Terms of an Arithmetic Progression**

*Theorem 2*

Suppose \(\left( {{u_n}} \right)\) is an arithmetic progression with a common difference \(d\). Let \({S_n} = {u_1} + {u_2} + \ldots + {u_n}\), then:

\({S_n} = \left( {{u_1} + {u_n}} \right)\dfrac{n}{2}\) or \({S_n} = n{u_1} + \dfrac{{n\left( {n – 1} \right)}}{2}d.

## SOME EXERCISES ON ARITHMETIC PROGRESSION

**Example 1.** Prove that the sequence $\left( {{u_n}} \right)$ with ${u_n} = 2020n – 2021$ is an arithmetic progression.

**Solution:**

We have ${u_{n + 1}} – {u_n} = 2020\left( {n + 1} \right) – 2021 – \left( {2020n – 2021} \right) = 2020.$

So, $\left( {{u_n}} \right)$ is an arithmetic progression with a common difference \(d = 2020.\)

**Example 2:** Given an arithmetic progression \(\left( {{u_n}} \right)\) with \({u_3} = 15\) and \(d = – 2\). Find \({u_n}.\)

**Solution:**

We have \(\left\{ \begin{array}{l}15 = {u_3} = {u_1} + 2d\\d = – 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1} = 19\\d = – 2\end{array} \right. \to {u_n} = {u_1} + \left( {n – 1} \right)d = – 2n + 21.

**Example 3:** An arithmetic progression has \(8\) terms. The first term is \(5\), and the eighth term is \(40\). What is the common difference \(d\) of this arithmetic progression?

**Solution:**

We have: \(\left\{ \begin{array}{l}{u_1} = 5\\40 = {u_8} = {u_1} + 7d\end{array} \right. \Rightarrow d = 5\)

**Example 4: **Given an arithmetic progression $\left( {{u_n}} \right)$ with \({u_1} + 2{u_5} = 0\) and \({S_4} = 14\). Calculate the first term \({u_1}\) and the common difference \(d\) of the arithmetic progression.

**Solution:**

We have \({u_1} + 2{u_5} = 0 \Leftrightarrow {u_1} + 2({u_1} + 4d) = 0 \Leftrightarrow 3{u_1} + 8d = 0\).

\({S_4} = 14 \Leftrightarrow \dfrac{{4(2{u_1} + 3d)}}{2} = 14 \Leftrightarrow 2{u_1} + 3d = 7\)

We have the system of equations \(\left\{ \begin{array}{l}3{u_1} + 8d = 0\\2{u_1} + 3d = 7\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1} = 8\\d = – 3\end{array} \right.\).

**Example 5:** Given an arithmetic progression \(\left( {{u_n}} \right)\) with \({u_1} = 4\) and \(d = – 5.\) Calculate the sum of the first \(100\) terms of the arithmetic progression.

**Solution:**

\({S_n} = n{u_1} + \frac{{n\left( {n – 1} \right)}}{2}d \Rightarrow {S_{100}} = 100{u_1} + \frac{{100.99}}{2}d = – 24350\)

**Example 6:** If the numbers \(5 + m;{\rm{ }}7 + 2m;{\rm{ }}17 + m\) arranged in order form an arithmetic progression, what is the value of \(m\)?

**Solution.**

The three numbers \(5 + m;{\rm{ }}7 + 2m;{\rm{ }}17 + m\) in order form an arithmetic progression, so

$ \left( {5 + m} \right) + \left( {17 + m} \right) = 2\left( {7 + 2m} \right) \Leftrightarrow m = 4$

**Example 7:** Prove that three positive numbers \(a,b,c\) arranged in order form an arithmetic progression if and only if the numbers \(\dfrac{1}{{\sqrt b + \sqrt c }},\dfrac{1}{{\sqrt c + \sqrt a }},\dfrac{1}{{\sqrt a + \sqrt b }}\) arranged in order form an arithmetic progression.

**Solution.**

We will prove by equivalent transformation.

Three numbers \(\dfrac{1}{{\sqrt b + \sqrt c }},\dfrac{1}{{\sqrt c + \sqrt a }},\dfrac{1}{{\sqrt a + \sqrt b }}\) form an arithmetic progression if and only if

\(\begin{array}{l}\dfrac{1}{{\sqrt c + \sqrt a }} – \dfrac{1}{{\sqrt b + \sqrt c }} = \dfrac{1}{{\sqrt a + \sqrt b }} – \dfrac{1}{{\sqrt c + \sqrt a }}\\ \Leftrightarrow \dfrac{{\sqrt b – \sqrt a }}{{(\sqrt c + \sqrt a )(\sqrt b + \sqrt c )}} = \dfrac{{\sqrt c – \sqrt b }}{{(\sqrt a + \sqrt b )(\sqrt c + \sqrt a )}}\end{array}\)

\( \Leftrightarrow (\sqrt b – \sqrt a )(\sqrt b + \sqrt a ) = (\sqrt c – \sqrt b )(\sqrt c + \sqrt b )\)

\( \Leftrightarrow \quad b – a = c – b \Leftrightarrow a,b,c\) arranged in order form an arithmetic progression.

**Example 8:** Given $a,b,c$ are three consecutive terms of an arithmetic progression, prove that

a) ${a^2} + 2bc = {c^2} + 2ab.$ b) ${a^2} + 8bc = {\left( {2b + c} \right)^2}.$

**Solution:**

Since $a,b,c$ are three consecutive terms of an arithmetic progression, $a + c = 2b \Leftrightarrow a = 2b – c.$

a) We have:

$\begin{array}{l}{a^2} – 2ab = {\left( {2b – c} \right)^2} – 2\left( {2b – c} \right)b\\ = 4{b^2} – 4bc + {c^2} – 4{b^2} + 2bc\\{\rm{ = }}{{\rm{c}}^2} – 2bc.\end{array}$

So, ${a^2} – 2ab = {c^2} – 2bc \Leftrightarrow {a^2} + 2bc = {c^2} + 2ab.$

b) We have ${a^2} + 8bc = {\left( {2b – c} \right)^2} + 8bc = 4{b^2} – 4bc + {c^2} + 8bc$

${\rm{ = 4}}{b^2} + 4bc + {c^2} = {\left( {2b + c} \right)^2}.$

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