### 等比數列

 A geometric sequence is a sequence having a common ratio between any term (except the first term) and its preceding terms. i.e. $\dfrac{{T}_{{2}}}{{T}_{{1}}}=\dfrac{{T}_{{3}}}{{T}_{{2}}}=$ … $=\dfrac{{T}_{{n}}}{{T}_{{{n}-{1}}}}=$ … , where ${T}_{{n}}$ is the general term.

 (a) If the first term is ${a}$ and the common ratio is ${r},$ then the general term is given by: ${T}_{{n}}={a}{r}^{{{n}-{1}}}$ , where ${n}$ is a positive integer. (b) Properties of geometric sequences (i) If ${T}_{{{n}-{1}}},{T}_{{n}}$ and ${T}_{{{n}+{1}}}$ are three consecutive terms of a geometric sequence, then ${\left({T}_{{n}}\right)}^{{2}}={T}_{{{n}-{1}}}\times{T}_{{{n}+{1}}}.$ (ii) If ${\left\lbrace{T}_{{1}}\right.}$ , ${T}_{{2}}$ , ${T}_{{3}}$ , … ${\rbrace}$ is a geometric sequence, then ${\left\lbrace{k}{T}_{{1}}\right.}$ , ${k}{T}_{{2}}$ , ${k}{T}_{{3}}$ , … ${\rbrace}$ is also a geometric sequence, where ${k}$ is a constant.

 Example Consider the geometric sequence ${48}$ , ${y}$ , ${12}$ , ... 

 (a) Find the possible value(s) of ${y}$ . (b) If all the terms of the sequence are positive, find  (i) its general term${T}{\left({n}\right)}$ , (ii) the value of ${k}$ such that the ${k}$th term is $\dfrac{{3}}{{64}}$ .

Solution

 (a) ∵ ${48}$ , ${y}$ , ${12}$ are in geometric sequence. ∴ ${y}^{{2}}$ $={48}\times{12}$ ${y}$ $={24}$ or ${y}=-{24}$ (b) (i) Let ${a}$ and ${r}$ be thr first term and the common ratio respectively. ∵   All the terms are positive. ∴   Take ${y}={24}$ . ${a}={48}$ and ${r}=\dfrac{{24}}{{48}}=\dfrac{{1}}{{2}}$ ∴ ${T}{\left({n}\right)}$ $={a}{r}^{{{n}-{1}}}$  $={48}{\left(\dfrac{{1}}{{2}}\right)}^{{{n}-{1}}}$  $={48}{\left(\dfrac{{1}}{{2}}\right)}^{{n}}{\left(\dfrac{{1}}{{2}}\right)}^{{-{1}}}$  $=\dfrac{{96}}{{{2}^{{n}}}}$ or ${3}{\left({2}\right)}^{{{5}-{n}}}$ (ii) From (b)(i), we have ${T}{\left({n}\right)}=\dfrac{{96}}{{{2}^{{n}}}}$ . ${T}{\left({k}\right)}$ $=\dfrac{{3}}{{64}}$ $\dfrac{{96}}{{2}^{{n}}}$ $=\dfrac{{3}}{{64}}$ ${2}^{{n}}$ $={2048}$ ${n}$ $={{\log}_{{2}}{2048}}$ ${{\log}_{{2}}{\left({2}^{{n}}\right)}}={{\log}_{{2}}{\left({2048}\right)}}$ ${n}$ $={11}$

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