### 等差數列

 An arithmetic sequence is a sequence having a common difference between any term (except the first term) and its preceding term. i.e. ${T}_{{2}}-{T}_{{1}}={T}_{{3}}-{T}_{{2}}={T}_{{n}}-{T}_{{{n}-{1}}}=$ … , where ${T}_{{n}}$ is the general term.

 (a) If the first term is ${a}$ and the common difference is ${d}$ , then the general term is given by: ${T}_{{n}}={a}+{\left({n}-{1}\right)}{d}$ , where ${n}$ is a positive integer. (b) Properties of arithmetic sequences (i) If ${T}_{{{n}-{1}}},{T}_{{n}}$ and ${T}_{{{n}+{1}}}$ are three consecutive terms of an arithmetic sequence, then ${T}_{{n}}=\dfrac{{{T}_{{{n}-{1}}}+{T}_{{{n}+{1}}}}}{{2}}$ . (ii) If ${\left\lbrace{T}_{{1}}\right.}$ , ${T}_{{2}}$ , ${T}_{{3}}$ , … ${\rbrace}$ is an arithmetic sequence, then ${\left\lbrace{k}{T}_{{1}}+{c}\right.}$ , ${k}{T}_{{2}}+{c}$ , ${k}{T}_{{3}}+{c}$ , … ${\rbrace}$ is also an arithmetic sequence, where ${k}$ and ${c}$ are constants.

 Example Consider the arithmetic sequence ${4}$ , ${10}$ , ${16}$ , ${22}$ , ... (a) Find the general term ${T}{\left({n}\right)}$ of the sequence. (b) Hence, show that the sequence with general term ${P}{\left({n}\right)}$ $={6}{n}-{7}$ is also an arithmetic sequence.

Solution

 (a) Let ${a}$ and ${d}$ be the first term and the common difference respectively. ${a}={4}$ and ${d}={10}-{4}={6}$ ∵ ${T}{\left({n}\right)}$ $={a}+{\left({n}-{1}\right)}{d}$  $={4}+{\left({n}-{1}\right)}{\left({6}\right)}$  $={6}{n}-{2}$ (b) ∵ ${P}{\left({n}\right)}$ $={6}{n}-{7}$  $={\left({6}{n}-{2}\right)}-{5}$  $={T}{\left({n}\right)}-{5}$ ,  and ${T}_{{1}}$ , ${T}_{{2}}$ , ${T}_{{3}}$ , … is an arithmetic sequence. ∴ ${P}{\left({1}\right)}$ , ${P}{\left({2}\right)}$ , ${P}{\left({3}\right)}$ , ... is also an arithmetic sequence.

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