Arithmetic and geometric sequences and their summations Examples


  • Examples
    The area of square \({A_1}{B_1}{C_1}{D_1}\) is \(1{\rm{ }}728{\rm{ c}}{{\rm{m}}^2}\).
    The mid-points of the sides of square \({A_1}{B_1}{C_1}{D_1}\) are joined to form square \({A_2}{B_2}{C_2}{D_2}\) and the subsequent squares are also obtained by this method.
    (See the figure)

    [GraphMissing Latex_Arithmetic and geometric sequences and their summations 02 Q28]
    (a) (i) Find the ratio of the area of square\({A_1}{B_1}{C_1}{D_1}\) to that of square \({A_2}{B_2}{C_2}{D_2}\).
         (ii) Express the area of square \({A_n}{B_n}{C_n}{D_n}\) in terms of n.
    (b) Find the area of square \({A_{10}}{B_{10}}{C_{10}}{D_{10}}\).
  • Solutions
    (a) (i) In \(\Delta {A_1}{A_2}{D_2}\),
    \(\begin{array}{1}{A_2}{D_2}^2 = {A_1}{A_2}^2 + {A_1}{D_2}^2\\ = {(\frac{{{A_1}{B_1}}}{2})^2} + {(\frac{{{A_1}{D_1}}}{2})^2}\\ = \frac{{{A_1}{D_1}^2}}{4} + \frac{{{A_1}{D_1}^2}}{4}\\ = \frac{{{A_1}{D_1}^2}}{2}(Pyth. theorem)\end{array}\)
    \(\frac{{{A_1}{D_1}^2}}{{{A_2}{D_2}^2}} = 2\)
    \(\begin{array}{1}\frac{{{\rm{A r e a o f s}}{\kern 1pt} {\rm{q}}{\kern 1pt} {\rm{u a r e }}{A_1}{B_1}{C_1}{D_1}}}{{{\rm{A r e a o f s}}{\kern 1pt} {\rm{q}}{\kern 1pt} {\rm{u a r e }}{A_2}{B_2}{C_2}{D_2}}} = \frac{{{A_1}{D_1}^2}}{{{A_2}{D_2}^2}}\\ = 2\end{array}\)
    ∴ Area of square\({A_1}{B_1}{C_1}{D_1}\) : Area of square\({A_2}{B_2}{C_2}{D_2} = \underline{\underline {2:1}} \)

    (ii) Let \({T_n}{\rm{ c}}{{\rm{m}}^2}\) be the area of square \({A_n}{B_n}{C_n}{D_n}\).
    \(\frac{{{A_1}{D_1}^2}}{{{A_2}{D_2}^2}} = 2\)
    Similarly,
    \(\frac{{{A_n}{D_n}^2}}{{{A_{n + 1}}{D_{n + 1}}^2}} = 2\)
    First term \({T_1} = 1{\rm{ }}728\)
    Common ratio \(\begin{array}{1}r = \frac{{{T_{n + 1}}}}{{{T_n}}}\\ = \frac{{{A_{n + 1}}{D_{n + 1}}^2}}{{{A_n}{D_n}^2}}\\ = \frac{1}{2}\end{array}\)
    General term \(\begin{array}{1}{T_n} = 1{\rm{ }}728{(\frac{1}{2})^{n - 1}}\\ = 27{(2)^6}{(\frac{1}{2})^{n - 1}}^{^{^{}}}\\ = 27{(\frac{1}{2})^{ - 6}}{(\frac{1}{2})^{n - 1}}^{^{^{}}}\\ = 27{(\frac{1}{2})^{n - 7}}^{^{^{}}}\end{array}\)
    ∴ Area of square \({A_n}{B_n}{C_n}{D_n}\) \( = {T_n}{\rm{ c}}{{\rm{m}}^{\rm{2}}}\)
    \( = \underline{\underline {27{{(\frac{1}{2})}^{n - 7}}{\rm{ c}}{{\rm{m}}^{\rm{2}}}}} \)


    (b) Area of square \({A_{10}}{B_{10}}{C_{10}}{D_{10}}\) \( = 27{(\frac{1}{2})^{10 - 7}}{\rm{ c}}{{\rm{m}}^{\rm{2}}}\)
    \( = \underline{\underline {\frac{{27}}{8}{\rm{ c}}{{\rm{m}}^{\rm{2}}}}} \)


  • Examples
    In the figure, straight lines \({L_{{\kern 1pt} 1}}\) and \({L_{{\kern 1pt} 2}}\) intersect at A.
    \({B_1},{\rm{ }}{B_2},{\rm{ }}{B_3},{\rm{ }}{B_4},{\rm{ }} \cdots \) are points on \({L_{{\kern 1pt} 1}}\), \({C_1},{\rm{ }}{C_2},\) \({C_3},{\rm{ }}{C_4},{\rm{ }} \cdots \) are points on \({L_{{\kern 1pt} 2}}\).
    It is given that \({B_n}{C_n}{C_{n + 1}}{D_n}\) is a square and \({B_{n + 1}}{D_n}{C_{n + 1}}\) is a straight line.
    [GraphMissing Latex_Arithmetic and geometric sequences and their summations 02 Q31]
    (a) Prove that \({B_1}{C_1},{\rm{ }}{B_2}{C_2},{\rm{ }}{B_3}{C_3},{\rm{ }}{B_4}{C_4},{\rm{ }} \cdots \) form a geometric sequence.
    (b) It is given that \({B_1}{C_1} = 1{\rm{ cm}}\) and \(\theta = 30^\circ \). Find the area of square \({B_8}{C_8}{C_9}{D_8}\). (Give your answer correct to 3 significant figures.)
  • Solutions
    (a) [GraphMissing Latex_Arithmetic and geometric sequences and their summations 02 Q31]
    In \(\Delta A{B_n}{C_n}\),
    \(\begin{array}{1}\tan \theta = \frac{{{B_n}{C_n}}}{{A{C_n}}}\\{B_n}{C_n} = A{C_n}\tan \theta \end{array}\)
    If \({B_n}{C_n} = A{C_n}\tan \theta \),
    then \({B_{n + 1}}{C_{n + 1}} = A{C_{n + 1}}\tan \theta \)
    ∴ \(\begin{array}{1}\frac{{{B_{n + 1}}{C_{n + 1}}}}{{{B_n}{C_n}}} = \frac{{A{C_{n + 1}}\tan \theta }}{{A{C_n}\tan \theta }}\\ = {\frac{{A{C_{n + 1}}}}{{A{C_n}}}^{}}\\ = {\frac{{A{C_n} + {C_n}{C_{n + 1}}}}{{A{C_n}}}^{}}\\ = 1 + {\frac{{{B_n}{C_n}}}{{A{C_n}}}^{}}\\ = 1 + \tan {\theta ^{}}\end{array}\)
    ∵ \(\frac{{{B_{n + 1}}{C_{n + 1}}}}{{{B_n}{C_n}}}\) is a constant.
    ∴ \({B_1}{C_1},{\rm{ }}{B_2}{C_2},{\rm{ }}{B_3}{C_3},{\rm{ }}{B_4}{C_4},{\rm{ }} \cdots \) form a geometric sequence.

    (b) Let \({T_n}{\rm{ c}}{{\rm{m}}^2}\) be the area of square \({B_n}{C_n}{C_{n + 1}}{D_n}\).
    \(\begin{array}{1}\frac{{{T_{n + 1}}}}{{{T_n}}} = \frac{{{B_{n + 1}}{C_{n + 1}}^2}}{{{B_n}{C_n}^2}}\\ = {(1 + \tan \theta )^2}\end{array}\)
    ∵ \(\frac{{{T_{n + 1}}}}{{{T_n}}}\) is a constant.
    ∴ \({T_1},{\rm{ }}{T_2},{\rm{ }}{T_3},{\rm{ }}{T_4},{\rm{ }} \cdots \) form a geometric sequence.
    First term\(\begin{array}{1}a = {1^2}\\ = 1\end{array}\)
    When \(\theta = 30^\circ \), \(\tan \theta = \frac{1}{{\sqrt 3 }}\)
    Common ratio\(r = {(1 + \frac{1}{{\sqrt 3 }})^2}\)
    ∴ \(\begin{array}{1}{T_n} = a{r^{n - 1}}\\ = (1){[{(1 + \frac{1}{{\sqrt 3 }})^2}]^{n - 1}}\\ = {(1 + \frac{1}{{\sqrt 3 }})^{2n - 2}}^{}\end{array}\)
    ∴ Area of square \({B_8}{C_8}{C_9}{D_8}\) \( = {T_8}{\rm{ c}}{{\rm{m}}^2}\)
    \( = {(1 + \frac{1}{{\sqrt 3 }})^{2(8) - 2}}{\rm{ c}}{{\rm{m}}^2}\)
    \( = \underline{\underline {590{\rm{ c}}{{\rm{m}}^2}}}\) (corr. to 3 sig. fig.)


  • Examples
    The sum of the 2nd and 3rd terms of a geometric sequence is 36, and the 4th term is 4 times the 2nd term.
    (a) Find the common ratio of the sequence.
    (b) Find the general term of the sequence.
    (c) It is given that the common ratio of the sequence is negative. If all negative terms are picked to form a new sequence,
        (i) find the general term of the new sequence.
        (ii) determine whether the new sequence is a geometric sequence.
  • Solutions
    (a) Let \({T_n} = a{r^{n - 1}}\) be the general term of the sequence,
    where a and r are the first term and the common ratio of the sequence respectively.
    ∵ \({T_4} = 4{T_2}\)
    ∴ \(\begin{array}{1}a{r^{4 - 1}} = 4(a{r^{2 - 1}})\\a{r^3} = 4ar\\{r^2} = 4\end{array}\)
    \(r = 2\) or -2


    (b) If \(r = 2\),
    ∵ \({T_2} + {T_3} = 36\)
    ∴ \(\begin{array}{1}a{(2)^{2 - 1}} + a{(2)^{3 - 1}} = 36\\a(2 + {2^2}) = 36\\6a = {36^{}}\\a = {6^{}}\end{array}\)
    General term\(\begin{array}{1}{T_n} = a{r^{n - 1}}\\ = 6{(2)^{n - 1}}\\ = 3 \times 2 \times {(2)^{n - 1}}\\ = \underline{\underline {3{{(2)}^n}}} \end{array}\)
    If \(r = - 2\),
    ∵ \({T_2} + {T_3} = 36\)
    ∴ \(\begin{array}{1}a{( - 2)^{2 - 1}} + a{( - 2)^{3 - 1}} = 36\\a[( - 2) + {( - 2)^2}] = {36^{}}\\2a = {36^{}}\\a = {18^{}}\end{array}\)
    General term \(\begin{array}{1}{T_n} = a{r^{n - 1}}\\ = 18{( - 2)^{n - 1}}\\ = - 9 \times ( - 2) \times {( - 2)^{n - 1}}\\ = \underline{\underline { - {\rm{ }}9{{( - 2)}^n}}} \end{array}\)


    (c) (i) ∵ \(r < 0\)
    ∴ \(r = - 2\) and \({T_n} = - {\rm{ }}9{( - 2)^n}\)
    ∵ The even-numbered terms of the sequence are the negative terms.
    ∴ Take \(n = 2m\), where m is a positive integer.
    General term \({U_m}\) of the new sequence \( = \underline{\underline { - {\rm{ }}9{{( - 2)}^{2m}}}} \)

    (ii) If \({U_m} = - 9{( - 2)^{2m}}\),
    then \(\begin{array}{1}{U_{m + 1}} = - 9{( - 2)^{2(m + 1)}}\\ = - 9{( - 2)^{2m + 2}}^{}\end{array}\)
    ∴ \(\begin{array}{1}\frac{{{U_{m + 1}}}}{{{U_m}}} = \frac{{ - {\rm{ }}9{{( - 2)}^{2m + 2}}}}{{ - {\rm{ }}9{{( - 2)}^{2m}}}}\\ = {( - 2)^2}^{}\\ = {4^{}}\end{array}\)
    ∵ \(\frac{{{U_{m + 1}}}}{{{U_m}}}\)is a constant.
    ∴ The new sequence is a geometric sequence.